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3 years ago

Write an equation of the line in slope intercept form through the points (0,-2) and (4,-2)

Mathematics

1 answer:

Ksivusya [100]3 years ago

3 0

Answer:

y=2x-2

Step-by-step explanation:

Slope=M so to find slope its just rise over run so for this equation you would do 0+4=4 2-2=0 so that would be 2/0 or simplified to 2 the b=y intercept so if you have a 0 as an x in either points the y of that point will always be equal to b

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Wesley Snipes earns a monthly salary of $1,685, plus a 8.5% commission on all sales over $2,000 each month. This month, his sale

kogti [31]

Answer:

none of theses choices. wesley makes $2185 this month

Step-by-step explanation:

5 0

3 years ago

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim x → (π/2)+

Otrada [13]

Answer:

The limit of the function at x approaches to $\frac{\pi}{2}$ is $-\infty$ .

Step-by-step explanation:

Consider the information:

$\lim_{x \to \frac{\pi}{2}}\frac{cos(x)}{1-sin(x)}$

If we try to find the value at $\frac{\pi}{2}$ we will obtained a $\frac{0}{0}$ form. this means that L'Hôpital's rule applies.

To apply the rule, take the derivative of the numerator:

$\frac{d}{dx}cos(x)=-sin(x)$

Now, take the derivative of the denominator:

$\frac{d}{dx}1-sin(x)=-cos(x)$

Therefore,

$\lim_{x \to \frac{\pi}{2}}\frac{-sin(x)}{-cos(x)}$

$\lim_{x \to \frac{\pi}{2}}\frac{sin(x)}{cos(x)}$

$\lim_{x \to \frac{\pi}{2}}tan(x)}$

Since, tangent function approaches -∞ as x approaches to $\frac{\pi}{2}$

, therefore, the original expression does the same thing.

Hence, the limit of the function at x approaches to $\frac{\pi}{2}$ is $-\infty$ .

5 0

3 years ago

What is the missing value (x,-3)

qwelly [4]

Answer:

There is no missing value, sir.

Step-by-step explanation:

you have to have a detailed explanation for the answer :)

3 0

3 years ago

How do I do this question?

mariarad [96]

Answer:

x = (y +2)/3

Step-by-step explanation:

y= 3x-2

y+2=3x

x=(y +2)/3

4 0

3 years ago

X (dy/dx) = (y-1/y+1)-y

notsponge [240]

F' $( \frac{y-1}{y+1})-y$ =
f' $( \frac{y-1}{y+1})$ -f'(y) [/tex]=
$\frac{f'(y-1)(y+1)-f'(y+1)(y-1)}{(y+1)^2} -1$ =
$\frac{(f'(y)-f'(1))(y+1)-(f'(y)+f'(1))(y-1)}{(y+1)^2} -1$ =
$\frac{(1-0)(y+1)-(1-0)(y-1)}{(y+1)^2} -1$ =
$\frac{(1)(y+1)-(1)(y-1)}{(y+1)^2} -1$ =
$\frac{(y+1)-(y-1)}{(y+1)^2} -1$ =
$\frac{y+1-y+1}{(y+1)^2} -1$ =
$\frac{2}{(y+1)^2} -1$

8 0

3 years ago