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Nat2105 [25]
3 years ago
8

On a coordinate plane, 2 cube root functions are shown. The first cube root function is a solid line and goes through (negative

8, 2), has an inflection point at (3.5, 4), and goes through (4, 5). The second cube root function is a dashed line and is represented by the equation f (x) = RootIndex 3 StartRoot x EndRoot. The function goes through (negative 8, negative 2), has an inflection point at (0, 0), and goes thorugh (8, 2).
Which equation is graphed along with the parent function?

g(x) = RootIndex 3 StartRoot x + 3 EndRoot + 4
g(x) = RootIndex 3 StartRoot x minus 3 EndRoot + 4
g(x) = RootIndex 3 StartRoot x minus 4 EndRoot + 3
g(x) = RootIndex 3 StartRoot x + 4 EndRoot + 3
Mathematics
2 answers:
muminat3 years ago
7 0

Answer:

it is b on ed 2021

Step-by-step explanation:

Margaret [11]3 years ago
6 0

Answer:

B: g(x) = RootIndex 3 StartRoot x minus 3 EndRoot + 4

Step-by-step explanation:

Got it right on Edge. Hope it helps!

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C=27π inches

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99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
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(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
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(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=
2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

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Step-by-step explanation:

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Step-by-step explanation:

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