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Free_Kalibri [48]

3 years ago

10

I NEED HELP!! How do you get the answer? I know it's A or B,because of the inequality

2 answers:

guapka [62]3 years ago

6 0

I believe that your answer would be A. y < -294

WARRIOR [948]3 years ago

4 0

I'm pretty sure the answer is A but i'm not 100% sure. :/

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The cylinder and the sphere below have the same radius and the same volume. What is the height of the cylinder?

lyudmila [28]

Answer:

Height of cylinder (h) = (4/3)R

Step-by-step explanation:

Given:

Radius of cylinder (r1) = R

Height of cylinder (h) = H

Radius of sphere (r2) = R

Volume of cylinder = volume of sphere

Find:

Height of cylinder (h) = H = ?

Computation:

$Volume\ of\ cylinder = volume\ of\ sphere\\\\ \pi (r1)^2h =\frac{4}{3} \pi (r2)^3\\\\\pi (R)^2h =\frac{4}{3} \pi (R)^3\\\\ h=\frac{4}{3} (R)$

Height of cylinder (h) = (4/3)R

4 0

3 years ago

What is 19 x 3.564239

Phoenix [80]

67.720541 multiply 19 by 3.564239

5 0

3 years ago

Read 2 more answers

The correlation in error terms that arises when the error terms at successive points in time are related is termed _____.

n200080 [17]

Answer:

auto correlation is the answer to the question

7 0

1 year ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

A study is undertaken to find out how much water, in gallons, is in Lake Michigan. Which type of data will the study analyze?

Drupady [299]

<span>A.
continuous data
gallons in a lake would be continuous data because it is not counted in integers but measured by how much volume is in the lake.
</span>

8 0

3 years ago