Question

1. the formula for the volume of a right cylinder is $v = \pi r^2 h$. in terms of $\pi$, what is the volume of a cylinder with $r = 4x$ and $h = 7x - 3$? teks a2.7(b), teks a2.1(a)(b)

a) $112\pi x^3 - 48\pi x$
b) $48\pi x^3 + 112\pi x^2$
c) $4\pi r x^2 + 7x - 3$
d) $112\pi x^3 - 48\pi x^2$

Explanation:

Step1: Substitute \( r = 4x \) and \( h = 7x - 3 \) into the volume formula \( V=\pi r^{2}h \)

First, calculate \( r^{2} \): \( (4x)^{2}=16x^{2} \)
Then, substitute into the formula: \( V = \pi\times(16x^{2})\times(7x - 3) \)

Step2: Distribute \( 16x^{2} \) over \( 7x - 3 \)

Using the distributive property \( a(b + c)=ab+ac \) (here \( a = 16x^{2}\pi \), \( b = 7x \), \( c=- 3 \)):
\( V=16x^{2}\pi\times7x-16x^{2}\pi\times3 \)
Calculate each term: \( 16\times7 = 112 \), so \( 16x^{2}\pi\times7x = 112\pi x^{3} \); \( 16\times3 = 48 \), so \( 16x^{2}\pi\times3 = 48\pi x^{2} \)
So \( V = 112\pi x^{3}-48\pi x^{2} \) Wait, there is a typo in my previous calculation, let's re - calculate:

Wait, no, let's do it again. \( r = 4x \), so \( r^{2}=(4x)^{2}=16x^{2} \), \( h = 7x-3 \)
\( V=\pi r^{2}h=\pi\times(4x)^{2}\times(7x - 3)=\pi\times16x^{2}\times(7x - 3) \)
Using the distributive law \( a(b - c)=ab - ac \) where \( a = 16x^{2}\pi \), \( b = 7x \), \( c = 3 \)
\( 16x^{2}\pi\times7x=112\pi x^{3} \)
\( 16x^{2}\pi\times3 = 48\pi x^{2} \)
So \( V=112\pi x^{3}-48\pi x^{2} \)? Wait, but in the options, option D is \( 112\pi x^{3}-48\pi x^{2} \) (I think there was a typo in the original problem's option D, maybe the exponent of the second term is 2). Let's check the options again. Option D is \( 112\pi x^{3}-48\pi x^{2} \) (assuming the original option D has a typo in the exponent of the second term, maybe it's \( x^{2} \) instead of \( x \)).

Wait, let's check the original problem's options again. The original option D is \( 112\pi x^{3}-48\pi x^{2} \) (probably a typo in the user's input, the second term's exponent is 2). So the correct answer is D.

Answer:

D. \( 112\pi x^{3}-48\pi x^{2} \)