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3 years ago

Helppp!!!!!

Mathematics

1 answer:

nydimaria [60]3 years ago

8 0

Answer:

true, false, true

Step-by-step explanation:

it only changes the direction

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Suppose that 21 inches of wire costs 63 cents.

Sergio [31]

<h2>Answer : 144 cents</h2>

Step by step explanation

Given: cost of 21 inches of wire costs 63 cents

Cost of 1 inch wire = 63/21 = 3 cents

Cost of 48 inches of wire = 48×3 = 144 cents

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4 0

2 years ago

Twice a number is decreased by 9, and this sum is multiplied by 4. The result is 8 less than 10 times the number. What is the nu

Sati [7]

4(2n - 9) = 10n - 8

8n - 36 = 10n - 8

8n - 8n - 36 = 10n - 8n - 8

-36 = 2n - 8

-36 + 8 = 2n - 8 + 8

-28 = 2n

-28/2 = 2n/2

-14 = n

7 0

3 years ago

Use the general slicing method to find the volume of the following solids. The solid whose base is the region bounded by the cur

Y_Kistochka [10]

Answer:

The volume is $V=\frac{64}{15}$

Step-by-step explanation:

The General Slicing Method is given by

Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is

$V=\int\limits^b_a {A(x)} \, dx$

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is

The key observation is that the width is the distance between the upper bounding curve $y = 2 - x^2$ and the lower bounding curve $y = x^2$

The width of each square is given by

$w=(2-x^2)-x^2=2-2x^2$

This means that the area of the square cross section at the point x is

$A(x)=(2-2x^2)^2$

The intersection points of the two bounding curves satisfy $2 - x^2=x^2$ , which has solutions x = ±1.

$2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}$

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

$V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}$