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3 years ago

This is a question about volume! Please help!!

Mathematics

1 answer:

Marina86 [1]3 years ago

7 0

Answer:

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

Step-by-step explanation:

$V=\frac{1}{3} \pi r^2h$

$\frac{1}{18} \pi =\frac{1}{3}\pi (\frac{2}{3} / 2 )^2 h$

$\frac{1}{18} =\frac{1}{3} (\frac{2}{6})^2 h$

$\frac{1}{18} =\frac{1}{3} (\frac{4}{36})h$

$\frac{1}{18} =\frac{4}{108} h$

$h=\frac{1}{18}*\frac{108}{4}$

$h=\frac{6}{4}$

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

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2 years ago

What type of number is -421

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2 years ago

Prove that the statement (ab)^n=a^n * b^n is true using mathematical induction.

mamaluj [8]

Answer:

see below

Step-by-step explanation:

(ab)^n=a^n * b^n

We need to show that it is true for n=1

assuming that it is true for n = k;

(ab)^n=a^n * b^n

( ab) ^1 = a^1 * b^1

ab = a * b

ab = ab

Then we need to show that it is true for n = ( k+1)

or (ab)^(k+1)=a^( k+1) * b^( k+1)

Starting with

(ab)^k=a^k * b^k given

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5 0

3 years ago

Read 2 more answers

Which of the following expressions is this one equivalent to?

schepotkina [342]

Answer: C. x-2

Step-by-step explanation:

We have the following expression:

$\frac{x^{4} -2x^{3} +x-2}{x^{3}+1}$

Factoring in the numerator:

$\frac{x(x^{3}+1)-2(x^{3}+1)}{x^{3}+1}$

Factoring again in the numerator with common factor $x^{3}+1$ :

$\frac{(x^{3}+1)(x-2)}{x^{3}+1}$

Simplifying:

$x-2$

Hence, the correct option is C. x-2

3 0

3 years ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

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3 years ago