write the expression as a decimal , 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000 =__
Answer:
6.986.
Step-by-step explanation:
6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000
We do the multiplications first ( according to PEMDAS):-
= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001
= 6 + 0.9 + 0.08 + 0006
= 6.9 + 0.086
= 6 986.
The value of the equation in the decimal form is A = 6.986
What is an Equation?
Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.
It demonstrates the equality of the relationship between the expressions printed on the left and right sides.
Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.
Given data ,
Let the equation be represented as A
Now , the value of A is
A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000
On simplifying the equation , we get
The value of 6 x 1 = 6
The value of 9 x 1/10 = 0.9
The value of 9 x 1/100 = 0.08
The value of 6 x 1/1000 = 0.006
So , substituting the values in the equation A , we get
A = 6 + 0.9 + 0.08 + 0.006
On simplifying the equation , we get
A = 6.986
Therefore , the value of A is 6.986
Hence , the value of the equation is 6.986
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The tree diagram below shows the possible combinations of juice and snack that can be offered at the school fair.
A tree diagram. Orange branches to popcorn and pretzels. Grape branches to popcorn and pretzels. Apple branches to popcorn and pretzels. Grapefruit branches to popcorn and pretzels.
How many different combinations are modeled by the diagram?
6
8
12
32
Answer:
B. 8Step-by-step explanation:
The combinations are:
Orange - 2 (with popcorn and pretzels)Grape - 2 (with popcorn and pretzels)Apple - 2 (with popcorn and pretzels)Grapefruit - 2 (with popcorn and pretzels)Total number of combinations:
4*2 = 8Correct choice is B
there are 8different combinations are modeled by the diagram.
Answer:
Solution given:
orange:2
grape:2
apple:2
grapefruit:2
no of term:4
now
total no. of combination ia 4*2=8
There are two rectangles Jared is examining. He knows the width of the first rectangle measures
2.48 cm, and the length is twice its width.
Jared also knows that the width of the second rectangle is equal to the length of the first rectangle,
and that the area of the second rectangle is 9.92. Given this information, find the length of the
second rectangle for Jared.
1st rectangle:
width: 2.48cm
length: 4.96
2nd rectangle:
width: 4.96 (equals to the length of the 1st rectangle)
area: 9.92
length: 9.92/4.96 = 2
It is known that the variance of a population equals 1,936. A random sample of 121 has been selected from the population. There is a .95 probability that the sample mean will provide a margin of error of _____. Group of answer choices 31.36 or less 1,936 or less 344.96 or less 7.84 or less
Answer:
Option d (7.84 or less) is the right alternative.
Step-by-step explanation:
Given:
[tex]\sigma^2=1936[/tex]
[tex]\sigma = \sqrt{1936}[/tex]
[tex]=44[/tex]
Random sample,
[tex]n = 121[/tex]
The level of significance,
= 0.95
or,
[tex](1-\alpha) = 0.95[/tex]
[tex]\alpha = 1-0.95[/tex]
[tex]Z_{\frac{\alpha}{2} } = 1.96[/tex]
hence,
The margin of error will be:
⇒ [tex]E = Z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )[/tex]
By putting the values, we get
[tex]=1.96(\frac{44}{\sqrt{121} } )[/tex]
[tex]=1.96(\frac{44}{11} )[/tex]
[tex]=1.96\times 4[/tex]
[tex]=7.84[/tex]
Beginning in January, a person plans to deposit $1 at the end of each month into an account earning
15% compounded monthly. Each year taxes must be paid on the interest earned during that year. Find
the interest earned during each year for the first 3 years.
Answer:
hi I am a Nepal
[tex] {233333}^{2332} [/tex]
What is the approximate length of arc s on the circle below? Use 3.14 for Pi. Round your answer to the nearest tenth.
-5.8 ft
-6.3 ft
-27.5 ft
-69.1 ft
9514 1404 393
Answer:
69.1 ft
Step-by-step explanation:
The diameter of the circle is 24 ft. The length of the arc is more than twice the diameter, so cannot be less than about 50 ft. The only reasonable choice is ...
69.1 ft
__
The circumference of the circle is ...
C = 2πr = 2(3.14)(12 ft) = 75.36 ft
The arc length of interest is 330° of the 360° circle, so is 330/360 = 11/12 times the circumference.
s = (11/12)(75.36 ft) = 69.08 ft ≈ 69.1 ft
Answer:D
Step-by-step explanation:
Select the correct answer.
What is the best way to describe a theme of this poem?
A.
The main purpose of having New Year's resolutions is to make people feel bad.
B.
The failures of the past should inspire people to accomplish more in the future.
OC.
By the end of the year, it is too late to make any changes to a person's life.
D.
People would accomplish their New Year's resolutions if they wrote them down.
B.The failures of the past should inspire people to accomplish more in the future.
A ball is dropped from a height of 14 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter an exact number.)
Answer:
Hello,
742/27 (ft)
Step-by-step explanation:
[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]
[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]
The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]
What is the total distance?
Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.
Here given that,
A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.
So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].
Then again it hits the ground and covers the distance [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is
[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]
Then it falls the same distance and goes back to the height
[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]
So, the total distance travelled is
[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]
We take the sum is twice because it goes back to the particular height and falls to the same distance.
[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]
Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]
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A ball is thrown in air and it's height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t)=−4t2+16t. When is the ball higher than 12 feet off the ground?
A. 3
B. 1
C. 1
D. 4
Hence the time that the ball will be height than 12 feet off the ground is 4secs
Given the expression for calculating the height in feet as;
h(t) = -4t²+16t
If the ball is higher than 12feet, h(t) > 12
Substituting h = 12 into the expression
-4t²+16t > 12
-4t²+16t - 12 > 0
4t²- 16t + 12 > 0
t²- 4t + 3 > 0
Factorize
(t²- 3t)-(t + 3) > 0
t(t-3)-1(t-3) > 0
(t-1)(t-3)>0
t > 1 and 3secs
Hence the time that the ball will be height than 12 feet off the ground is 4secs
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Geometry help I don’t get this stuff at all
Answer:
The last option
V = (-1.5,3)
other options dont lie where V is exact
V is only Exact at (-1.5,3)
Which of the following describes a positive correlation?
As the number of hours spent on homework increases, the tests scores increase.
As the number of apples eaten per year increases, the number of times visiting the doctor every year remains the same.
As the number of times going to bed early increases, the number of times waking up late decreases.
The amount of time a team spent practicing increases, the number of games lost in a season decreases.
THIS IS A MULTIPLE CHOICE QUESTION
Answer:
First Choice: As the number of hours spent on homework increases, the tests scores increase.
Step-by-step explanation:
The definition of a positive correlation is a relationship between two given variables, in which both variables are moving in the same direction. This can mean when one variable increases and the other variable increases, too, or one variable decreases and the other decreases as well.
The first choice is a positive correlation because both variables are changing (increasing) in the same direction. As you spend more time on homework, you're likely to get a higher test score.
The second choice cannot be a positive correlation because only one variable is having some kind of change (increasing). The doctor visits amount remains the same, so we can call this a zero-correlation relationship because the number of apples eaten yearly doesn't affect the amount of doctor visits. An apple a day keeps the doctor a way is just a proverb, not to be taken literally.
The third choice cannot be a positive correlation because the two variables are going different directions. Even though the number of times going to bed early is increasing, the number of times waking up late decreases, which is not moving in the same direction as the other variable.
The fourth choice cannot be a positive correlation because, similarly to the third choice, the two variables are going different directions. One variable is increasing, which is the amount of practice time. Meanwhile, the other variable is decreasing (going in the opposite direction), which is the number of games lost in a season.
A cell phone company charges a monthly fee of $18 plus five cents for each call. A
customer's total cell phone bill this month is $50.50. Use n to represent the number of
calls.
Answer:
650 calls
Step-by-step explanation:
so since you have 18$ per month plus 5 cents per call you would do
18+0.5n(n represent the number of calls)= the total fee of $50.50 cents.
thus,now you need to figure out how much the phone calls were without the monthly fee so you would do:
50.50-18=32.50
so 32.50 is the price of all the phone calls
then you divide 32.50 by 0.05 which equals to 650
meaning that n=650
hope I helped!
Find the minimum sample size needed to be 99% confident that the sample's variance is within 30% of the population's variance.
The Minimum sample size table is attached below
Answer:
[tex]X=173[/tex]
Step-by-step explanation:
From the question we are told that:
Confidence Interval [tex]CI=99\%[/tex]
Variance [tex]\sigma^2=30\%[/tex]
Generally going through the table the
Minimum sample size is
[tex]X=173[/tex]
help i need help with math help if u can
Previous studies suggest that use of nicotine-replacement therapies and antidepressants can help people stop smoking. The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. The target for quitting smoking was the 8th day of the experiment.
In this experiment researchers randomly assigned smokers to treatments. Of the 162 smokers taking a placebo, 28 stopped smoking by the 8th day. Of the 272 smokers taking only the antidepressant buproprion, 82 stopped smoking by the 8th day.
Calculate the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment). (The standard error is about 0.0407. Use critical value z = 2.576.)
( ), ( )
Round your answer to three decimal places. Put lower bound in the first box and upper bound in the second box.
Using the z-distribution, it is found that the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).
What is a t-distribution confidence interval?The confidence interval is:
[tex]\overline{x} \pm zs[/tex]
In which:
[tex]\overline{x}[/tex] is the sample mean.z is the critical value.s is the standard error.In this problem, we are given that z = 2.576, s = 0.0407. The sample mean is the difference of the proportions, hence:
[tex]\overline{x} = \frac{28}{162} - \frac{82}{272} = -0.129[/tex]
Then, the bounds of the interval are given by:
[tex]\overline{x} - zs = -0.129 - 2.576(0.0407) = -0.234[/tex]
[tex]\overline{x} + zs = -0.129 + 2.576(0.0407) = -0.024[/tex]
The 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).
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A researcher is interested in whether there is a significant difference between the mean age of marriage across three racial groups. Using the data provided below, conduct an F-test to determine whether you believe there is an association between race and average age at marriage.
Race N Mean
Black 113 25.39
White 904 22.99
Other 144 23.87
All Groups 1,161 23.33
Answer:
The P-value is < significance value ( 0.05 ) hence we reject the Null hypothesis ( i.e. There is an association between the race and average age at marriage )
Step-by-step explanation:
Conducting an F-test to determine association between race and average age at marriage
step 1 : State the hypothesis
H0 : ц1 = ц2 = ц3
Ha : ц1 ≠ ц2 ≠ ц3
step 2 : determine the mean square between
Given mean value of all groups = 23.33
SS btw = 113*(25.39 - 23.33)² + 904*(22.99 - 23.33)² + 144*(23.89 - 23.33)^2 = 113(4.2436) + 904(0.1156) + 144(0.3136)
= 629.1876
hence: df btw = 3 - 1 = 2
df total = 1161 - 1 = 1160
df within = 1160 - 2 = 1158
SS within = 36.87*1158 = 42695.46
Therefore the MS between = 629.19 / 2 = 314.60
The F-ratio = 314.59 / 36.87 = 8.53
using the values for Btw the P-value = 0.00021
The P-value is < significance value ( 0.05 ) hence we reject the Null hypothesis ( i.e. There is an association between the race and average age at marriage )
0.003 is 1/10 of
Please help I need this for homework !!!!!!!!!!!!
Answer:
0.03
Step-by-step explanation:
Question 4 (2 marks)
Justin works 14 hours at a normal pay rate of $24.80 per hour and 5 hours of overtime at
time and a half. How much should he be paid?
I
809 words
LE
English (Australia)
Answer:
554.7
Step-by-step explanation:
The pay=25.8*14+(25.8)*5*1.5=554.7
The three sides of a triangle are n, 3n+3, and 3n−1. If the perimeter of the triangle is 72m, what is the length of each side?
Answer: 10m, 33m, and 29m
Step-by-step explanation:
n + 3n+3 + 3n-1 = 72m
7n+2=72m
7n = 72-2
n = 70/7
n = 10
I am struggling and I would be so happy if any of you helped me. Can someone help me with the last two red boxes please? The rest of the question is for reference to help solve the problem. Thank you for your time!
Answer:
I think you can go with:
The margin of error is equal to half the width of the entire confidence interval.
so try .74 ± = [ .724 , .756] as the confidence interval
Step-by-step explanation:
Which number would be rounded UP to the nearest ten but DOWN to the nearest hundred?
A. 232
B. 238
C. 262
D. 268
Answer:
B
Step-by-step explanation:
a special window in the shape of a rectangle with semicircles at each end is to be constructed so that the outside perimeter is 100 feet. find the dimensions of the rectangle tha tmaximizes the total area of the window
Answer:
The dimensions of the rectangle are length 25 feet and width 15.92 feet
Step-by-step explanation:
Let L be the length of the rectangle and w be the width.
The area of the rectangular part of the shape is Lw while the area of the two semi-circular ends which have a diameter which equals the width of the rectangle is 2 × πw²/8 = πw²/4. The area of each semi-circle is πw²/4 ÷ 2 = πw²/8
So, the area of the shape A = Lw + πw²/4.
The perimeter of the shape, P equals the length of the semi-circular sides plus twice its length. The length of a semi-circular side is πw/2. So, both sides is 2 × πw/2 = πw
P = πw + 2L
Since the perimeter, P = 100 feet, we have
πw + 2L = 100
From this L = (100 - πw)/2
Substituting L into A, we have
A = Lw + πw²/4.
A = [(100 - πw)/2]w + πw²/4.
A = 50w - πw²/2 + πw²/4.
A = 50w - πw²/2
Now differentiating A with respect to w and equating it to zero to find the value of w which maximizes A.
So
dA/dw = d[50w - πw²/2]/dw
dA/dw = 50 - πw
50 - πw = 0
πw = 50
w = 50/π = 15.92 feet
differentiating A twice to get d²A/dw² = - π indicating that w = 50/π is a value at which A is maximum since d²A/dw² < 0.
So, substituting w = 50/π into L, we have
L = (100 - πw)/2
L = 50 - π(50/π)/2
L = 50 - 50/2
L = 50 - 25
L = 25 feet
So, the dimensions of the rectangle are length 25 feet and width 15.92 feet
Let a=⟨1,−4,2⟩ and b=⟨−5,−5,−2⟩. Compute:
a+b=⟨ ,, ⟩
a−b=⟨ ,,⟩
2a=⟨ ,,⟩
3a+4b=⟨ ,, ⟩
|a|=
Answer:
a + b = ⟨-4, -9, 0⟩
a - b = ⟨6, 1, 4⟩
2a = ⟨2, -8, 4⟩
3a + 4b = ⟨-17, -32, -2⟩
|a| = √21
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightPre-Calculus
Vectors
OperationsScalars[Magnitude] ||v|| = √(x² + y² + z²)Step-by-step explanation:
Adding and subtracting vectors are follow the similar pattern of normal order of operations:
a + b = ⟨1 - 5, -4 - 5, 2 - 2⟩ = ⟨-4, -9, 0⟩
a - b = ⟨1 + 5, -4 + 5, 2 + 2⟩ = ⟨6, 1, 4⟩
Scalar multiplication multiplies each component:
2a = ⟨2(1), 2(-4), 2(2)⟩ = ⟨2, -8, 4⟩
Remember to multiply in the scalar before doing basic operations:
3a + 4b = ⟨3(1), 3(-4), 3(2)⟩ + ⟨4(-5), 4(-5), 4(-2)⟩ = ⟨3, -12, 6⟩ + ⟨-20, -20, -8⟩ = ⟨-17, -32, -2⟩
Absolute values surrounding a vector signifies magnitude of a vector. Follow the formula:
|a| = √[1² + (-4)² + 2²] = √21
To calculate the volume of a chemical produced in a day a chemical manufacturing company uses the following formula below:
[tex]V(x)=[C_1(x)+C_2(x)](H(x))[/tex]
where represents the number of units produced. This means two chemicals are added together to make a new chemical and the resulting chemical is multiplied by the expression for the holding container with respect to the number of units produced. The equations for the two chemicals added together with respect to the number of unit produced are given below:
[tex]C_1(x)=\frac{x}{x+1} , C_2(x)=\frac{2}{x-3}[/tex]
The equation for the holding container with respect to the number of unit produced is given below:
[tex]H(x)=\frac{x^3-9x}{x}[/tex]
a. What rational expression do you get when you combine the two chemicals?
b. What is the simplified equation of ?
c. What would the volume be if 50, 100, or 1000 units are produced in a day?
d. The company needs a volume of 3000 How many units would need to be produced in a day?
Answer:
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
[tex]V(50) = 2548.17[/tex] [tex]V(100) = 10098.10[/tex] [tex]V(1000) = 999201.78[/tex]
[tex]x = 54.78[/tex]
Step-by-step explanation:
Given
[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]
[tex]C_1(x) = \frac{x}{x+1}[/tex]
[tex]C_1(x) = \frac{2}{x-3}[/tex]
[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]
Solving (a): Expression for V(x)
We have:
[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]
Substitute known values
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
Solving (b): Simplify V(x)
We have:
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
Solve the expression in bracket
[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
Factor out x
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]
Express as difference of two squares
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]
Cancel out x - 3
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Solving (c): V(50), V(100), V(1000)
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Substitute 50 for x
[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]
[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]
[tex]V(50) = 2548.17[/tex]
Substitute 100 for x
[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]
[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]
[tex]V(100) = 10098.10[/tex]
Substitute 1000 for x
[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]
[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]
[tex]V(1000) = 999201.78[/tex]
Solving (d): V(x) = 3000, find x
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Cross multiply
[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]
Equate to 0
[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]
Open brackets
[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]
Collect like terms
[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]
[tex]x^3 + x^2 -3001x -2994 = 0[/tex]
Solve using graphs (see attachment)
[tex]x = -54.783[/tex] or
[tex]x = -0.998[/tex] or
[tex]x = 54.78[/tex]
x can't be negative. So:
[tex]x = 54.78[/tex]
write your answer as an integer or as a decimal rounded to the nearest tenth
Answer:
123456-6-&55674
Step-by-step explanation:
rdcfvvzxv.
dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see
recall see
The Susan B. Anthony dollar has a radius of 0.52 inches. Find the area of one side of the coin to the nearest
hundredth.
Answer:
0.85 in²
Step-by-step explanation:
really ? you need help with that ? you could not find the formula for the area of a circle on the internet and type it into your calculator ? I can't do anything else here.
a circle area is
A = pi×r²
r being the radius.
and pi being, well, pi (3.1415....)
r = 0.52 in
so,
A = pi×0.52² = pi×0.2704 = 0.849486654... in²
the area of one side of the coin is 0.85 in²
Which is the solution to-x/2<-4
A x<-8
B x2-8
C x <8
D x 8
Answer:
A.x<-8
Step-by-step explanation:
=1/2x<−4
=2*(1/2x)< (2)*(-4)
= x<-8
Michelle would like to know how much of her loan payments will go toward interest. She has a $124,500 loan with a 5.9% interest rate that is compounded monthly. The loan has a term of 10 years. Calculate the total amount of interest that Michelle will pay over the course of the loan.
9514 1404 393
Answer:
$40,615.20
Step-by-step explanation:
The amortization formula will tell you Michelle's monthly payment.
A = P(r/12)/(1 -(1 +r/12)^(-12t)) . . . . loan value P at interest rate r for t years
A = $124,500(0.059/12)/(1 -(1 +0.059/12)^(-12·10)) ≈ $1375.96
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The total of Michelle's 120 monthly payments is ...
12 × $1375.96 = $165,115.20
This amount pays both principal and interest, so the amount of interest she pays is ...
$165,115.20 -124,500 = $40,615.20
Michelle will pay $40,615.20 in interest over the course of the loan.
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A calculator or spreadsheet can figure this quickly.
y varies directly as the cube of x. When x = 3, then y = 7. Find y when x = 4.
Answer:
[tex]y \: \alpha \: {x}^{3} \ \\ y \: = k {x}^{3} \\ where \: y = 7 \: and \:x = 3 \\ y = k {x}^{3} \\ 7 = k ( {3)}^{3} \\ 7 = 27k \\ k = \frac{7}{27} \\ \\ so \: \: y = \frac{7}{27} {x}^{3} \\ \\ y = \frac{7}{27} {4}^{3} \\ y = \frac{448}{27} [/tex]
the required value of y at x = 4 is 16.64.
Given that,
y varies directly as the cube of x. When x = 3, then y = 7.To determine the y when x = 4.
Proportionality is defined as between two or more sets of values, and how these values are related to each other in the sense that are they directly proportional or inversely proportional to each other.
Here,
y is directly proportional to the cube of x i.e
y ∝ x³
y = kx³ - - - - - (1)
where k is proportionality constant,
At x = 3 y = 7
7 = k (3)³
7 / 27 = k
k = 0.26
Put k in equation 1
y = 0.26 x³
Now at x = 4
y = 0.26 * 4³
y = 0.26 * 64
y = 16.64
Thus, the required value of y at x = 4 is 16.64.
Learn more about proportionality here: https://brainly.com/question/22620356
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[infinity]
Substitute y(x)= Σ 2 anx^n and the Maclaurin series for 6 sin3x into y' - 2xy = 6 sin 3x and equate the coefficients of like powers of x on both sides of the equation to n= 0. Find the first four nonzero terms in a power series expansion about x = 0 of a general
n=0
solution to the differential equation.
У(Ñ)= ___________
Recall that
[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]
Differentiating the power series series for y(x) gives the series for y'(x) :
[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]
Now, replace everything in the DE with the corresponding power series:
[tex]y'-2xy = 6\sin(3x) \implies[/tex]
[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]
The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.
Split up both series on the left into even- and odd-indexed series:
[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]
[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]
Next, we want to condense the even and odd series:
• Even:
[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]
• Odd:
[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]
Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].
The even series vanishes, so that
[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]
for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find
[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]
[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]
and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].
This leaves us with the odd series,
[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]
[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]
We have
[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]
[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]
[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]
[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]
So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then
[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]
[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]
[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]
and so the first four terms of series solution to the DE would be
[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]
