Factor 3x2 +6x- 24 Please help me

Readme [11.4K]

Surface Area of the figure is 1208 square centimeters
a=20
b=13
c=12
d=5
e=8
Top face:
A1=a×e=20×8
ae=160
Bottom face:
A2(a+2d)×e=(20+2×5)×8=(20+10)×8=30×8
(a+2d)e=240
Front face:
Rectangle
a×c=20×12=240
Triangle:
12c×d=12×12×5=6×5=30
A3Rectangle +2 triangles
240+2×30=240+60=300
(a+d)c=300
Slant face:
A4=b×e=13×8=104
be=104
Total surface area
A=A1+A2+2(A3+A4)
A1=160
A2=240
A3=300
A4=104
Thus,
A=160+240+2(300+104)
A=400+2(404)
A=400+808=1208
Surface Area of the figure is 1208 square centimeters

8 0

3 years ago

Of

ollegr [7]

The answer is 25 I think sorry if it’s wrong :)

3 0

2 years ago

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Compare the slopes of the linear functions f(x) and g(x) and choose the answer that best describes them.

Tcecarenko [31]

Answer:

The slope of f(x) is equal to the slope of g(x).

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope of the function f(x)

we have the points

(0,-1) and (3,1)

substitute in the formula

$m=\frac{1+1}{3-0}$

$m_1=\frac{2}{3}$

step 2

Find the slope of the function g(x)

take two points of the given table

(0,2) and (3,4)

substitute in the formula

$m=\frac{4-2}{3-0}$

$m_2=\frac{2}{3}$

step 3

Compare the slopes

$m_1=m_2$

therefore

The slope of f(x) is equal to the slope of g(x).

7 0

3 years ago

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22 hours. A

SIZIF [17.4K]

Answer:

100/8 = 12.5 (s)

Step-by-step explanation:

hmm bro

7 0

3 years ago

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Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves

Vadim26 [7]

The expression on the left side describes a parabola. Factorize it to determine where it crosses the y-axis (i.e. the line x = 0) :

-3y² + 9y - 6 = -3 (y² - 3y + 2)

… = -3 (y - 1) (y - 2) = 0

⇒ y = 1 or y = 2

Also, complete the square to determine the vertex of the parabola:

-3y² + 9y - 6 = -3 (y² - 3y) - 6

… = -3 (y² - 3y + 9/4 - 9/4) - 6

… = -3 (y² - 2•3/2 y + (3/2)²) + 27/4 - 6

… = -3 (y - 3/2)² + 3/4

⇒ vertex at (x, y) = (3/4, 3/2)

I've attached a sketch of the curve along with one of the shells that make up the solid. For some value of x in the interval 0 ≤ x ≤ 3/4, each cylindrical shell has

radius = x

height = y⁺ - y⁻

where y⁺ refers to the half of the parabola above the line y = 3/2, and y⁻ is the lower half. These halves are functions of x that we obtain from its equation by solving for y :

x = -3y² + 9y - 6

x = -3 (y - 3/2)² + 3/4

x - 3/4 = -3 (y - 3/2)²

-x/3 + 1/4 = (y - 3/2)²

± √(1/4 - x/3) = y - 3/2

y = 3/2 ± √(1/4 - x/3)

y⁺ and y⁻ are the solutions with the positive and negative square roots, respectively, so each shell has height

(3/2 + √(1/4 - x/3)) - (3/2 - √(1/4 - x/3)) = 2 √(1/4 - x/3)

Now set up the integral and compute the volume.

$\displaystyle 2\pi \int_{x=0}^{x=3/4} 2x \sqrt{\frac14 - \frac x3} \, dx$

Substitute u = 1/4 - x/3, so x = 3/4 - 3u and dx = -3 du.

$\displaystyle 2\pi \int_{u=1/4-0/3}^{u=1/4-(3/4)/3} 2\left(\frac34 - 3u\right) \sqrt{u} \left(-3 \, du\right)$

$\displaystyle -12\pi \int_{u=1/4}^{u=0} \left(\frac34 - 3u\right) \sqrt{u} \, du$

$\displaystyle 12\pi \int_{u=0}^{u=1/4} \left(\frac34 u^{1/2} - 3u^{3/2}\right) \, du$

$\displaystyle 12\pi \left(\frac34\cdot\frac23 u^{3/2} - 3\cdot\frac25u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 u^{3/2} - \frac65u^{5/2}\right) \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 \left(\frac14\right)^{3/2} - \frac65\left(\frac14\right)^{5/2}\right) - 12\pi (0 - 0)$

$\displaystyle 12\pi \left(\frac1{16} - \frac3{80}\right) = \frac{12\pi}{40} = \boxed{\frac{3\pi}{10}}$

6 0

2 years ago