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Archy [21]
3 years ago
12

Which statement correctly compares the two functions given below? f(x) =-32x-17 g(x) =-8x+12 A. Function f(x) decreases at a rat

e 4 times faster than function g(x) B. Function g(x) decreases at a rate 24 times faster than function f(x) C. Function g(x) decreases at a rate 4 times faster than function f(x) D. Function f(x) decreases at a rate 24 times faster than function g(x)
​
Mathematics
1 answer:
Leokris [45]3 years ago
6 0

Answer: A

Step-by-step explanation:

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total surface area  = 298.45105 m2

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top surface area = 78.53975 m2

bottom surface area = 78.53975 m2

Step-by-step explanation:

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Is 32.971 bigger then 39.217
malfutka [58]
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No

Explanation 32.971 is smaller than 39.217
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A cone is placed inside a cylinder as shown. The radius of the cone is half the radius of the cylinder. The height of the cone i
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volume of the cylinder divided by 2 is the volume of the cone

Step-by-step explanation:

4 0
3 years ago
Find the volume of the given solid. Bounded by the cylinder y2 + z2 = 16 and the planes x = 2y, x = 0, z = 0 in the first octant
erastovalidia [21]

Answer:

\mathbf{  \dfrac{128}{3}}

Step-by-step explanation:

Given that:

y^2 + z^2 = 16; \\ \\ where; \ x = 2y , x =0, z= 0

Replace \ z = 0  \ \text{in the given equation ;} y^2+z^2=16

Then;

y^2 = 16 \\ \\  y = \sqrt{16} \\ \\ y = \pm 4

\text{So; in 1st  octant \ limits of y }= 0 \to 4 \ \text{and for x }= 0 \to 2y

∴

\text{Volume of solid: (V) = }\int ^{4}_{y=0} \int ^{2y}_{x=0} \sqrt{16-y^2} \ dxdy \\ \\ = \int^4_{y=0} \sqrt{16 -y^2} \ (x)^{2y} _o \ dy  \\ \\ =  \int^4_{y=0}\sqrt{16 -y^2} (2y -0) \ dy \\ \\  V = \int^{4}_{y=0} \ 2y \sqrt{16 - y^2} \ dy

Now, let:\\\\ 16 - y^2 = u  \\ \\  -2ydy = du  \\ \\  2ydy = -du \\ \\ Hence, when \  y = 0; u = 16 \\ \\  and \\ \\  when \ y =  4 \ then \ u = 0

∴

V = \int ^4 _{y=0} \ 2y \sqrt{16 -y^2 } \ dy  \\ \\  = \int ^0_{16} - \sqrt{u}\ du  \\ \\  = - \int^0_{16} \ u^{^{\dfrac{1}{2}}} \ du  \\ \\  = \Bigg [\dfrac{u^{^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1 } \Bigg]^0_{16} \\ \\ =\Bigg[ -\dfrac{u^{\dfrac{3}{2}}}{\dfrac{3}{2}}^0_{16} \Bigg] \\ \\

= - \dfrac{2}{3}\Big( u^{\dfrac{3}{2}}\Big)^0_{16} \\ \\ = - \dfrac{2}{3} \Big(0^{^\dfrac{3}{2}}}- 16^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(0 - (4^2)^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(-(4)^{3}}\Big )

= - \dfrac{2}{3} \Big(64\Big )

∴

\mathbf{V = \dfrac{128}{3}}

5 0
3 years ago
7+7-2x4{44-2+8)??????
Svetach [21]

Answer:

1577

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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