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2 years ago

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What are the variables in the expression 2w + 11t + 4y

1 answer:

Taya2010 [7]2 years ago

5 0

They are T, W, and Y

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Please help ASAP!

marusya05 [52]

Answer:

Not a function

Step-by-step explanation:

This is not a function because there are different outputs for the same input.

8 0

2 years ago

Read 2 more answers

PLEASE HELP ME OUT THIS IS DUE TODAY AND I AM STRESSED

spin [16.1K]

Answer:

look up answer key realy would help

Step-by-step explanation:

8 0

2 years ago

Determine the standard variation of the data below. (1, 2, 3, 4, 5)

Ne4ueva [31]

Calculate for the mean/ average of the given numbers:

μ = (1 + 2 + 3 + 4 + 5) / 5 = 3

Then, we calculate for the summation of the squares of differences of these numbers from the mean, S

S = (1 - 3)² + (2 - 3)² + (3 - 3)² + (4 - 3)² + (5 - 3)²
S = 10

Divide this summation by the number of items and take the square root of the result to get the standard deviation.

SD = sqrt (10 / 5) = sqrt 2
SD = 1.41

Thus, the standard deviation of the given is equal to 1.41.

8 0

3 years ago

Read 2 more answers

One plus the product of a number y and three

slamgirl [31]

(yX3)+1 because product means multipy

8 0

2 years ago

find the volume of the solid formed by revolving the region bounded by the graphs of y = 4x - x^2 and f(x) = x^2 from [0,2] abou

Neko [114]

Answer:

$v = \frac{32\pi}{3}$

or

$v=33.52$

Step-by-step explanation:

Given

$f(x) = 4x - x^2$

$g(x) = x^2$

$[a,b] = [0,2]$

Required

The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

$\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx$

Integrate

$v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx$

$v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx$

Substitute values for a, b, f(x) and g(x)

$v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx$

Evaluate the exponents

$v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx$

Simplify like terms

$v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx$

Factor out 8

$v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx$

Integrate

$v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0$

$v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0$

Substitute 2 and 0 for x, respectively

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])$

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])$

$v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]$

$v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]$

Take LCM

$v = 8\pi [ \frac{16*4- 16 * 3}{12}]$

$v = 8\pi [ \frac{64- 48}{12}]$

$v = 8\pi * \frac{16}{12}$

Simplify

$v = 8\pi * \frac{4}{3}$

$v = \frac{32\pi}{3}$

or

$v=\frac{32}{3} * \frac{22}{7}$

$v=\frac{32*22}{3*7}$

$v=\frac{704}{21}$

$v=33.52$

8 0

2 years ago