Does anyone know how to do this ? please help i don’t know how to . all three if you can :(

What are the independent and dependent variables in the relationship?

Sindrei [870]

w is the width and it is the independent variable. You can pick any positive whole number you want (within reason of course; you can't go to infinity or go beyond some set boundary). Whatever you picked for w, the expression 2w-5 will be dependent on it. So the length is dependent on the width.

For instance, if the width is w = 10 feet, then 2w-5 = 2*10-5 = 20-5 = 15 feet is the length. The choice of 10 feet for the width directly affects the length being 15 feet.

5 0

3 years ago

I increase a number by 24% the answer is 6014 what number did I start with

miskamm [114]

Answer:

1443.36

Step-by-step explanation:

6014*.24=1443.36

Hope it helps

8 0

3 years ago

The value of the correlation coefficient is always in the range a. 0 to 1 b. 1 to 1 c. 1 to 0

dlinn [17]

Answer:

-1 to 1

Step-by-step explanation:

The correlation Coefficient gives the degree of relationship between two variables while also giving an hint about the type of relationship between them, (positive or negative). Correlation Coefficient could take any value bwuqwen - 1 and 1. With values closer to - 1 or 1 indicating a strong relationship between the variables, Coefficient of 0 means no relationship and a negative value means negative relationship while a positive value plies a positive relationship.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

Find the missing side lengths leave your answer as a racials simplest form

Aleksandr-060686 [28]

Answer:

m= $7\sqrt3$

n=7

Step-by-step explanation:

Hi there!

We are given a right triangle (notice the 90°) angle, the measure of one of the acute angles as 60°, and the measure of the hypotenuse (the side OPPOSITE from the 90 degree angle) as 14

We need to find the lengths of m and n

Firstly, let's find the measure of the other acute angle

The acute angles in a right triangle are complementary, meaning they add up to 90 degrees

Let's make the measure of the unknown acute angle x

So x+60°=90°

Subtract 60 from both sides

x=30°

So the measure of the other acute angle is 30 degrees

This makes the right triangle a special kind of right triangle, a 30°-60°-90° triangle

In a 30°-60°-90° triangle, if the length of the hypotenuse is a, then the length of the leg (the side that makes up the right angle) opposite from the 30 degree angle is $\frac{a}{2}$ , and the leg opposite from the 60 degree angle is $\frac{a\sqrt3}{2}$

In this case, a=14, n= $\frac{a}{2}$ , and m= $\frac{a\sqrt3}{2}$

Now substitute the value of a into the formulas to find n and m to find the lengths of those sides

So that means that n= $\frac{14}{2}$ , which is equal to 7

And m= $\frac{14\sqrt3}{2}$ , which simplified, is equal to $7\sqrt3$

Hope this helps!

7 0

3 years ago