Answer:
5/9
Step-by-step explanation:
What we have in this question are four white balls and 5 red balls.
This is the sample space
[(4w,0R) (4w,1R)(4w,2R)(4w,3R)(4w,4R)(0w,5R)(1w,5R)(2w,5R)
So we have 9 possible sets of events
The event number with the last ball being white = 5
Probability of the last ball drawn being white = 5/9
= 0.56
Find the sum of ∑3/k=0 k^2
Answer:
[tex]14[/tex]
Step-by-step explanation:
Given
[tex]\displaystyle \sum_{k=0}^3k^2[/tex]
Let's break down each part. The input at the bottom, in this case [tex]k=0[/tex], is assigning an index [tex]k[/tex] at a value of [tex]0[/tex]. This is the value we should start with when substituting into our equation.
The number at the top, in this case 3, indicates the index we should stop at, inclusive (meaning we finish substituting that index and then stop). The equation on the right, in this case [tex]k^2[/tex], is the equation we will substitute each value in. After we substitute our starting index, we'll continue substituting indexes until we reach the last index, then add up each of the outputs produced.
Since [tex]k=0[/tex] is our starting index, start by substituting this into [tex]k^2[/tex]:
[tex]0^2=0[/tex]
Now continue with [tex]k=1[/tex]:
[tex]1^1=1[/tex]
Repeat until we get to the ending index, [tex]k=3[/tex]. Remember to still use [tex]k=3[/tex] before stopping!
Substituting [tex]k=2[/tex]:
[tex]2^2=4[/tex]
Substituting [tex]k=3[/tex]:
[tex]3^2=9[/tex]
Since 3 is the index we end at, we stop here. Now we will add up each of the outputs:
[tex]0+1+4+9=\boxed{14}[/tex]
Therefore, our answer is:
[tex]\displaystyle \sum_{k=0}^3k^2=0+1+4+9=\boxed{14}[/tex]
Answer:
14
Step-by-step explanation:
∑3/k=0 k^2
Let k=0
0^2 =0
Let k = 1
1^2 =1
Let k =2
2^2 = 4
Let k = 3
3^2 = 9
0+1+4+9 = 14
why infinity ( ) can’t be included in an inequality?
Answer:
Step-by-step explanation:
Because then the value on the other side will be unbounded by the infinity sign while expressing the answers on a number line.
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use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x
First check the characteristic solution: the characteristic equation for this DE is
r ² - 3r + 2 = (r - 2) (r - 1) = 0
with roots r = 2 and r = 1, so the characteristic solution is
y (char.) = C₁ exp(2x) + C₂ exp(x)
For the ansatz particular solution, we might first try
y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)
where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).
However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :
y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)
Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.
y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)
… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)
y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)
… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)
Substituting every relevant expression and simplifying reduces the equation to
(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)
… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]
… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]
= 2x + (1 + 2x) exp(x) + 4 exp(3x)
… … …
2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)
= 2x + (1 + 2x) exp(x) + 4 exp(3x)
Then, equating coefficients of corresponding terms on both sides, we have the system of equations,
x : 2a = 2
1 : -3a + 2b = 0
exp(x) : 2c - d = 1
x exp(x) : -2c = 2
exp(3x) : 2e = 4
Solving the system gives
a = 1, b = 3/2, c = -1, d = -3, e = 2
Then the general solution to the DE is
y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)
Which of the following statements are true?
Answer:
last one
Step-by-step explanation:
they both whole
inveres laplace transform (3s-14)/s^2-4s+8
Complete the square in the denominator.
[tex]s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4[/tex]
Rewrite the given transform as
[tex]\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}[/tex]
Now take the inverse transform:
[tex]L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}[/tex]
A manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs of solvent. Suppose the amount of solvent in each drum is normally distributed with a mean of 101.8 pounds and a standard deviation of 3.76 pounds.
Required:
a. What is the probability that a drum meets the guarantee? Give your answer to four decimal places.
b. What would the standard deviation need to be so that the probability a drum meets the guarantee is 0.99?
Answer:
The answer is "0.6368 and 0.773".
Step-by-step explanation:
The manufacturer of organic compounds guarantees that its clients have at least 100 lbs. of solvent in every fluid drum they deliver. [tex]X\ is\ N(101.8, 3.76)\\\\P(X>100) =P(Z> \frac{100-101.8}{3.76}=P(Z>-0.47))[/tex]
For point a:
Therefore the Probability =0.6368
For point b:
[tex]P(Z\geq \frac{100-101.8}{\sigma})=0.99\\\\P(Z\geq \frac{-1.8}{\sigma})=0.99\\\\1-P(Z< \frac{-1.8}{\sigma})=0.99\\\\P(Z< \frac{-1.8}{\sigma})=0.01\\\\z-value =0.01\\\\area=-2.33\\\\ \frac{-1.8}{\sigma}=-2.33\\\\ \sigma= \frac{-1.8}{-2.33}=0.773[/tex]
Please help!
What is the pattern,
Y-interception
And equation
Answer: y=1x+1
Step-by-step explanation:
y=1x+3
that should be it
Which expression defines the given series for seven terms?
–4 + (–5) + (–6) + . . .
Answer: -n+(-n-1)
Step-by-step explanation:
Expression will be -n + (-1)
Series
-4 +(-5)+(-6)+(-7)+(-8)+(-9)+(-10)+(-11)+(-12)+(-13) and so on
Here number -n has + (-n-1) being added to it
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13
R
S
12
What's the length of QR?
A) 1
B) 17.7
C) 6.7
OD) 5
Answer:
5
Step-by-step explanation:
This is a right triangle, so we can use the Pythagorean theorem
a^2+b^2 = c^2
where a and b are the legs and c is the hypotenuse
QR^2 + 12^2 = 13^2
QR^2 +144 =169
QR^2 = 169-144
QR^2 =25
Take the square root of each side
QR = sqrt(25)
QR =5
Please answer this question
The breadth of a rectangular garden is 2/3 of its legth. If its perimeter is 40cm, find its dimensions.
Answer:
12; 8
Step-by-step explanation:
length-x
breadth-2/3x
2(x+2/3 x)=40
2×5/3x=40
10/3x=40
x=40÷10/3
x=40×3/10
x=12 (cm) length
2/3×12=8 (cm) breadth
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the wall. A. At what rate is the ladder sliding away from the wall when the base of the ladder is 10 feet from the wall
[tex]7.2\:\text{ft/s}[/tex]
Step-by-step explanation:
We can apply the Pythagorean theorem here:
[tex]26^2 = x^2 + y^2\:\:\:\:\:\:\:\:\:(1)[/tex]
where x is the distance of the ladder base from the wall and y is the distance of the ladder top from the ground. Taking the time derivative of the expression above, we get
[tex]0 = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}[/tex]
Solving for [tex]\frac{dx}{dt},[/tex] we get
[tex]\dfrac{dx}{dt} = -\dfrac{y}{x}\dfrac{dy}{dt}[/tex]
We can replace y by rearranging Eqn(1) such that
[tex]y = \sqrt{26^2 - x^2}[/tex]
Therefore,
[tex]\dfrac{dx}{dt} = - \dfrac{\sqrt{26^2 - x^2}}{x}\dfrac{dy}{dt}[/tex]
Since y is decreasing as the ladder is being lowered, we will assign a negative sign to [tex]\frac{dy}{dt}[/tex]. Hence,
[tex]\dfrac{dx}{dt} = - \dfrac{\sqrt{26^2 - (10)^2}}{10}(-3\:\text{ft/min})[/tex]
[tex]\:\:\:\:\:\:\:= 7.2\:\text{ft/min}[/tex]
In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 55.4 inches, and standard deviation of 4.1 inches.
A) What is the probability that a randomly chosen child has a height of less than 61.25 inches?
Answer= (Round your answer to 4 decimal places.)
B) What is the probability that a randomly chosen child has a height of more than 46.5 inches?
Answer= (Round your answer to 4 decimal places.)
(A)
P(X < 61.25) = P((X - 55.4)/4.1 < (61.25 - 55.4)/4.1)
… ≈ P(Z ≤ 0.1427)
… ≈ 0.5567
(B)
P(X > 46.5) = P((X - 55.4)/4.1 > (46.5 - 55.4)/4.1)
… ≈ P(Z > -2.1707)
… ≈ 1 - P(Z ≤ -2.1707)
… ≈ 0.9850
help with 1 b please. using ln.
Answer:
[tex]\displaystyle \frac{dy}{dx} = \frac{1}{(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]
General Formulas and Concepts:
Pre-Algebra
Equality PropertiesAlgebra I
Terms/CoefficientsFactoringExponential Rule [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]Algebra II
Natural logarithms ln and Euler's number eLogarithmic Property [Exponential]: [tex]\displaystyle log(a^b) = b \cdot log(a)[/tex]Calculus
Differentiation
DerivativesDerivative NotationImplicit DifferentiationDerivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
*Note:
You can simply just use the Quotient and Chain Rule to find the derivative instead of using ln.
Step 1: Define
Identify
[tex]\displaystyle y = \sqrt{\frac{x}{2 - x}}[/tex]
Step 2: Rewrite
[Function] Exponential Rule [Root Rewrite]: [tex]\displaystyle y = \bigg( \frac{x}{2 - x} \bigg)^\bigg{\frac{1}{2}}[/tex][Equality Property] ln both sides: [tex]\displaystyle lny = ln \bigg[ \bigg( \frac{x}{2 - x} \bigg)^\bigg{\frac{1}{2}} \bigg][/tex]Logarithmic Property [Exponential]: [tex]\displaystyle lny = \frac{1}{2}ln \bigg( \frac{x}{2 - x} \bigg)[/tex]Step 3: Differentiate
Implicit Differentiation: [tex]\displaystyle \frac{dy}{dx}[lny] = \frac{dy}{dx} \bigg[ \frac{1}{2}ln \bigg( \frac{x}{2 - x} \bigg) \bigg][/tex]Logarithmic Differentiation [Derivative Rule - Chain Rule]: [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{1}{2} \bigg( \frac{1}{\frac{x}{2 - x}} \bigg) \frac{dy}{dx} \bigg[ \frac{x}{2 - x} \bigg][/tex]Chain Rule [Basic Power Rule]: [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{1}{2} \bigg( \frac{1}{\frac{x}{2 - x}} \bigg) \bigg[ \frac{2}{(x - 2)^2} \bigg][/tex]Simplify: [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{-1}{x(x - 2)}[/tex]Isolate [tex]\displaystyle \frac{dy}{dx}[/tex]: [tex]\displaystyle \frac{dy}{dx} = \frac{-y}{x(x - 2)}[/tex]Substitute in y [Derivative]: [tex]\displaystyle \frac{dy}{dx} = \frac{-\sqrt{\frac{x}{2 - x}}}{x(x - 2)}[/tex]Rationalize: [tex]\displaystyle \frac{dy}{dx} = \frac{-\frac{x}{2 - x}}{x(x - 2)\sqrt{\frac{x}{2 - x}}}[/tex]Rewrite: [tex]\displaystyle \frac{dy}{dx} = \frac{-x}{x(x - 2)(2 - x)\sqrt{\frac{x}{2 - x}}}[/tex]Factor: [tex]\displaystyle \frac{dy}{dx} = \frac{-x}{-x(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{1}{(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Book: College Calculus 10e
If per unit variable cost of a product is Rs.8 and fixed cost is Rs 5000 and it is sold for Rs 15 per unit, profit in 1000 units is.......
a.. rs 7000
b. rs 2000
c. rs 25000
d. rs 0
Answer:
a.. rs 7000
Because 15×1000=15000 it is SP when selling 1000units in the rate of Rs 15/unit& 8×1000=8000 this is cp when buying 1000 units in the rate of Rs 8/unit.
So,by formula of profit,
Rs (15000-8000)=Rs7000
Write a quadratic equation having the given numbers as solutions. -7 and -5
The quadratic equation is ___ =0.
Answer:
x²+12x+35
Step-by-step explanation:
in factored form it would just be
(x+7)(x+5)=0
expand this
x²+12x+35=0
Any number that CAN be divided by 2 without having remainder is considered an _______ number
Step-by-step explanation:
Any number that can be divided by 2 without having remainder is considered an even number.
I hope it helped U
stay safe stay happy
Form a union for the following sets.
M = {1, 2, 4, 8)
N = (2,5,8)
Answer:
Step-by-step explanation:
When you are asked to find the union of sets you find numbers that are present in both sets.
So a number that appears in both the sets of M and N are 2 and 8.
So M U N = { 2,8} where U is the symbol for union.
What's 14,124 ÷ 44 ?
[tex]14124 \div 44[/tex]
Answer:
321
Step-by-step explanation:
We are given a weighted coin (with one side heads, one side tails), and we want to estimate the unknown probability pp that it will land heads. We flip the coin 1000 times and it happens to land heads 406 times. Give answers in decimal form, rounded to four decimal places (or more). We estimate the chance this coin will land on heads to
Answer:
0.4060
Step-by-step explanation:
To calculate the sample proportion, phat, we take the ratio of the number of preferred outcome to the total number of trials ;
Phat = number of times coin lands on head (preferred outcome), x / total number of trials (total coin flips), n
x = 406
n = 1000
Phat = x / n = 406/ 1000 = 0.4060
The estimate of the chance that this coin will land on heads is 0.406
Probability is the likelihood or chance that an event will occur.Probability = Expected outcome/Total outcomeIf a coin is flipped 1000 times, the total outcomes will 1000
If it landed on the head 406 times, the expected outcome will be 406.
Pr(the coin lands on the head) = 406/1000
Pr(the coin lands on the head) = 0.406
Hence the estimate of the chance that this coin will land on heads is 0.406
Learn more on probability here: https://brainly.com/question/14192140
How to find joint and combined variation?
Step-by-step explanation:
Z will stay the same since the 2 will divide into 1
The following 3 points are on a parabola defining the edge of a ski.
(-4, 1), (-2, 0.94), (0,1)
The general form for the equation of a parabola is:
Ax^2 + Bx + C= y
Required:
a. Use the x- and y-values of 1 of the points to build a linear equation with 3 variables: A, B, and C.
b. Record your equation here. Repeat this process with 1 of the other 2 points to build a 2nd linear equation.
c. Record your equation here. Repeat this process with the other point to build a 3rd equation.
d. Record your equation here. Build a matrix equation that represents this system of equations.
e. Record your matrix equation here. Use a graphing calculator or other graphing utility to find the inverse of the coefficient matrix.
f. Record your result here. Use the inverse matrix to solve the system of equations. Record the equation of the parabola here.
a. The linear equation for the first point (-4,1) is 16A-4B+C=1
b. The linear equation for the second point (-2, 0.94) is 4A-2B+C=0.94
c. The linear equation for the third point (0,1) is 0A+0B+C=1
d. The matrix equation looks like this:
[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]
e. The inverse of the coefficient matrix looks like this:
[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]
f. The equation of the parabola is: [tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]
a. In order to build a linear equation from the given points, we need to substitute them into the general form of the equation.
Let's take the first point (-4,1). When substituting it into the general form of the quadratic equation we end up with:
[tex](-4)^{2}A+(-4)B+C=1[/tex]
which yields:
[tex]16A-4B+C=1[/tex]
b. Let's take the second point (-2,0.94). When substituting it into the general form of the quadratic equation we end up with:
[tex](-2)^{2}A+(-2)B+C=0.94[/tex]
which yields:
[tex]4A-2B+C=0.94[/tex]
c. Let's take the third point (0,1). When substituting it into the general form of the quadratic equation we end up with:
[tex](0)^{2}A+(0)B+C=1[/tex]
which yields:
[tex]0A+0B+C=1[/tex]
d. A matrix equation consists on three matrices. The first matrix contains the coefficients (this is the numbers on the left side of the linear equations). Make sure to write them in the right order, this is, the numbers next to the A's should go on the first column, the numbers next to the B's should go on the second column and the numbers next to the C's should go on the third column.
The equations are the following:
16A-4B+C=1
4A-2B+C=0.94
0A+0B+C=1
So the coefficient matrix looks like this:
[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right][/tex]
Next we have the matrix that has the variables, in this case our variables are the letters A, B and C. So the matrix looks like this:
[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right][/tex]
and finally the matrix with the answers to the equations, in this case 1, 0.94 and 1:
[tex]\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]
so if we put it all together we end up with the following matrix equation:
[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]
e. When inputing the coefficient matrix in our graphing calculator we end up with the following inverse matrix:
[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]
Inputing matrices and calculating their inverses depends on the model of a calculator you are using. You can refer to the user's manual on how to do that.
f. Our matrix equation has the following general form:
AX=B
where:
A=Coefficient matrix
X=Variables matrix
B= Answers matrix
In order to solve this type of equations, we can make use of the inverse of the coefficient matrix to end up with an equation that looks like this:
[tex]X=A^{-1}B[/tex]
Be careful with the order in which you are doing the multiplication, if A and B change places, then the multiplication will not work and you will not get the answer you need. So when solving this equation we get:
[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]*\left[\begin{array}{c}1\\\frac{47}{50}\\1\end{array}\right][/tex]
(Notice that I changed 0.94 for the fraction 47/50 you can get this number by dividing 94/100 and simplifying the fraction)
So, in order to do the multiplication, we need to multiply each row of the coefficient matrix by the answer matrix and add the results. Like this:
[tex]\frac{1}{8}*1+(-\frac{1}{4})(\frac{47}{50})+\frac{1}{8}*1[/tex]
[tex]\frac{1}{8}-\frac{47}{200}+\frac{1}{8}=\frac{3}{200}[/tex]
So the first number for the answer matrix is [tex]\frac{3}{200}[/tex]
[tex]\frac{1}{4}*1+(-1)(\frac{47}{50})+\frac{3}{4}*1[/tex]
[tex]\frac{1}{4}-\frac{47}{50}+\frac{3}{4}=\frac{3}{50}[/tex]
So the second number for the answer matrix is [tex]\frac{3}{50}[/tex]
[tex]0*1+0(\frac{47}{50})+1*1[/tex]
[tex]0+0+1=1[/tex]
So the third number for the answer matrix is 1
In the end, the matrix equation has the following answer.
[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}\frac{3}{200}\\\frac{3}{50}\\1\end{array}\right][/tex]
which means that:
[tex]A=\frac{3}{200}[/tex]
[tex]B=\frac{3}{50}[/tex]
and C=1
so, when substituting these answers in the general form of the equation of the parabola we get:
[tex]Ax^{2}+Bx+C=y[/tex]
[tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]
For further information, you can go to the following link:
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4. Jack started packing the box shown with 1-centimeter cubes.
- cubic centimeter
Select all the true statements below.
A. Jack needs to add 2 more layers to fill the box.
B. Jack packed 16 cubes into the bottom of the box.
The box is 8 centimeters long.
The box is 3 centimeters high.
E.) The volume of the box is 32 cubic centimeters.
F. The volume of the box is 16 centimeters.
Answer:
if I am not mistakedn the answe id e
if sin150=1/2 then find sin75
Answer:
0.966
Step-by-step explanation:
When typed into a calculator, sin75 = -0.3877816354
Upon converting to degrees, the full answer is 0.96592582628
Again need help with these ones I don’t understand and they have to show work
Which statement is true about the ratios of squares to
cicles in the tables? PLS HURRY!!!!
Answer:
show us a screenshot or image
or type it out, copy paste
Step-by-step explanation:
Answer as soon as you can. a. 162 comes just after b. What comes just before 182. lies in between 99 and 101. c.
Answer:
a. 161
b. 181
c. 100
Step-by-step explanation:
a. 162 comes just after 161 (160, 161, 162, 163...)
b. 181 comes just before 182 (180, 181, 182, 183...)
c. 100 is between 99 and 101 (98, 99, 100, 101, 102...)
please help! 50 points!
Answer:
a) forming a bell
b) 5
c) 4.7
d) mean
is the correct answer
pls mark me as brainliest
12) Find the angles between 0o and 360o where sec θ = −3.8637 . Round to the nearest 10th of a degree:
Please show all work
9514 1404 393
Answer:
105.0°, 255.0°
Step-by-step explanation:
Many calculators do not have a secant function, so the cosine relation must be used.
sec(θ) = -3.8637
1/cos(θ) = -3.8637
cos(θ) = -1/3.8637
θ = arccos(-1/3.8637) ≈ 105.000013°
The secant and cosine functions are symmetrical about the line θ = 180°, so the other solution in the desired range is ...
θ = 360° -105.0° = 255.0°
The angles of interest are θ = 105.0° and θ = 255.0°.
Use the information below to complete the problem: p(x)=1/x+1 and q(x)=1/x-1 Perform the operation and show that it results in another rational expression. p(x) + q(x)
Answer:
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Answer:
[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]
The result is indeed another rational expression.
Step-by-step explanation:
We are given the two functions:
[tex]\displaystyle p(x) = \frac{1}{x+1}\text{ and } q(x) = \frac{1}{x-1}[/tex]
And we want to perform the operation:
[tex]\displaystyle p(x) + q(x)[/tex]
And show that the result is another rational expression.
Add:
[tex]\displaystyle = \frac{1}{x+1} + \frac{1}{x-1}[/tex]
To combine the fractions, we will need a common denominator. So, we can multiply the first fraction by (x - 1) and the second by (x + 1):
[tex]\displaystyle = \frac{1}{x+1}\left(\frac{x-1}{x-1}\right) + \frac{1}{x-1}\left(\frac{x+1}{x+1}\right)[/tex]
Simplify:
[tex]=\displaystyle \frac{x-1}{(x+1)(x-1)} + \frac{x+1}{(x+1)(x-1)}[/tex]
Add:
[tex]\displaystyle = \frac{(x-1)+(x+1)}{(x+1)(x-1)}[/tex]
Simplify. Hence:
[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]
The result is indeed another rational expression.
