Answer:
B is the answer
Step-by-step explanation:
hope it helps
If a, b, c are in A.P. show that
a (b + c)/bc,b(c + a) /ca, c(a-b )/bc
are in A.P.
Answer:
Step-by-step explanation:
[tex]\frac{a(b+c)}{bc} ,\frac{b(c+a)}{ca} ,\frac{c(a+b)}{ab} ~are~in~A.P.\\if~\frac{ab+ca}{bc} ,\frac{bc+ab}{ca} ,\frac{ca+bc}{ab} ~are~in~A.P.\\add~1~to~each~term\\if~\frac{ab+ca}{bc} +1,\frac{bc+ab}{ca} +1,\frac{ca+bc}{ab} +1~are~in~A.P.\\if~\frac{ab+ca+bc}{bc} ,\frac{bc+ab+ca}{ca} ,\frac{ca+bc+ab\\}{ab} ~are~in~A.P.\\\\divide~each~by~ab+bc+ca\\if~\frac{1}{bc} ,\frac{1}{ca} ,\frac{1}{ab} ~are ~in~A.P.\\if~\frac{a}{abc} ,\frac{b}{abc} ,\frac{c}{abc} ~are~in~A.P.\\if~a,b,c~are~in~A.P.\\which~is~true.[/tex]
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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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
Enter the degree of the polynomial below.
6x + 9x + 3x – 4410 - 9x5 – 5x6
A. 9
B. 10
c. 6.
OD. 4
Answer:
the answer is d
Step-by-step explanation:
[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]
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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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
Enter the ratio as a fraction in lowest terms
6 minutes to 30 minutes.
6 minutes / 30 minutes
Divide the top and bottom by 6.
1 minute / 5 minutes
Fraction in lowest terms: 1/5
Hope this helps!
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:
I will give brainly.
How do you determine if a slope is positive or negative?
You have to find the slope .
How?
Take 2points
(x1,y1)(x2,y2)Slope formula[tex]\\ \rm\Rrightarrow \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Step-by-step explanation:
What the Slope Means A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y also decreases. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
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
PLZZZ HELP
This is due in 15 mins
I need 5
But I already have 4
So one more
Answer:
The hottest month for the northern hemisphere is August.
The hottest month for the southern hemisphere is January and February (these top two might be the opposite)
It's globally warmer during the months of June July and August
During april and november, the southern hemisphere and northern hemisphere are the same, or very close.
During July and August the southern and northern hemispheres have the largest difference in temperature
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]
Which graph is a function?
Answer:
B
Step-by-step explanation:
A function is a relation in which each input, x, has only one output, y.
There are two ways to determine if a relation is a function:
1. If each x-input has only one, unique y-output, then it's a function. If some x-inputs share the same y-outputs, it's not a function.
2. Vertical Line Test on Graphs:
To determine whether y is a function of x, when given a graph of relation, use the following criterion: if every vertical line you can draw goes though only 1 point, the relation can be a function. If you can draw a vertical line that goes though more than 1 point, the relation cannot be a function.
Since we're given a graph relation, let's test both of the answers out.
If I were to draw a vertical line in a specific place on the first graph, I'd be hitting more than one point in the coordinate plane.
If I were to draw a vertical line in a specific place on the second graph, I'd only be hitting one point in the coordinate plane.
Therefore, choice B is a function.
Answer please answer!!
I need the answer asap
Answer:
35 cm
Step-by-step explanation:
is the correct answer
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
HELP!!
Consider the polynomial
Answer:
1. coefficient of 3rd term = 1
2. constant term= 0
The coefficient of the third term is 1 while the constant term is 0 for the given expression.
What is an expression?An expression is a combination of some mathematical symbol such that an arithmetic operator and variable such that all are constrained and create an equation.
For example 3x +5y
As per the given polynomial,
(1/2)a⁴ + 3a³ + a
Here a is a variable.
(1)
The third term is a and its coefficient is 1 as (1)a.
(2)
All terms have variable "a" thus none of the terms is constant so the constant term is 0.
Hence "For the following statement, the constant term has a coefficient of 0 and the third term has a coefficient of 1".
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Let f(x) = e ^3x/5x − 2. Find f'(0).
Answer:
Step-by-step explanation:
Our friend asking what the actual function is has a point. I completed this under the assumption that what we have is:
[tex]f(x)=\frac{e^{3x}}{5x-2}[/tex] and used the quotient rule to find the derivative, as follows:
[tex]f'(x)=\frac{e^{3x}(5)-[(5x-2)(3e^{3x})]}{(5x-2)^2}[/tex] and simplifying a bit:
[tex]f'(x)=\frac{5e^{3x}-[15xe^{3x}-6e^{3x}]}{(5x-2)^2}[/tex]and a bit more to:
[tex]f'(x)=\frac{5e^{3x}-15xe^{3x}+6e^{3x}}{(5x-2)^2}[/tex] and combining like terms:
[tex]f'(x)=\frac{11e^{3x}-15xe^{3x}}{(5x-2)^2}[/tex] and factor out the GFC in the numerator to get:
[tex]f'(x)=\frac{e^{3x}(11-15x)}{(5x-2)^2}[/tex] That's the derivative simplified. If we want f'(0), we sub in 0's for the x's in there and get the value of the derivative at x = 0:
[tex]f'(0)=\frac{e^0(11-15(0))}{(5(0)-2)^2}[/tex] which simplifies to
[tex]f'(0)=\frac{11}{4}[/tex] which translates to
The slope of the function is 11/4 at the point (0, -1/2)
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
__
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.
__
A calculator or spreadsheet can figure this quickly.
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.
Please answer in detail
Answer:
y=5x-1 I think because the snd option doesn't make sense but you should try y =5x-1
Cho f là hàm chẵn, g là hàm lẻ. Tính giá trị của (g∘f)(−4,7), biết g(5,9)=7,9 và f(4,7)=5,9.
Step-by-step explanation:
yah language mujhe samajh mein nahin a rahi hai kya karu aapki is bataiye
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:
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
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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solve for x please help (show work)
Answer:
x = -1
Step-by-step explanation:
1/3(6x-3)+2(x-1) = -7
Distribute
2x-1 +2x-2 = -7
Combine like terms
4x -3 = -7
Add 3 to each side
4x-3+3 = -7+3
4x = -4
Divide by 4
4x/4 = -4/4
x = -1
[tex] \frak{ \frac{1}{3}(6x - 3) + 2(x - 1) = - 7}[/tex]
[tex]\twoheadrightarrow \frak{2x - 1 + 2x - 2 = - 7}[/tex]
[tex]\twoheadrightarrow \frak{4x - 3 = - 7}[/tex]
[tex] \twoheadrightarrow \frak{4x = - 7 + 3}[/tex]
[tex]\twoheadrightarrow \frak{4x = - 4}[/tex]
[tex]\star \: \underline{ \boxed{ \frak \green{{x = - 1}}}}[/tex]
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
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
Instructions: Determine whether the following polygons are
similar. If yes, type in the similarity statement and scale factor. If
no, type 'None' in the blanks.
Answer:
None
Step-by-step explanation:
The given angles aren't equal which is needed for the polygon to be similar
No, the following polygons are not similar.
Used the concept of a similar figure that states,
In terms of Maths, when two figures have the same shape but their sizes are different, then such figures are called similar figures.
Given that,
Two polygons EFGH and JKLM are shown in the image.
Now the corresponding sides of both figures are,
EF = 27
JK = 63
And, EH = 27
JM = 63
Hence, the ratio of corresponding sides is,
EF/JK = 27/63
= 9/21
= 3/7
EH/JM = 27/63
= 3/7
So their corresponding sides are equal in ratio.
But their corresponding angles are not the same.
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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]