Solve x^2 +11x = -10

I need help on this it is a fraction subtract another fraction which is negative thx to whoever helps 5/7-(-1/7)

Sav [38]

Answer:

$\frac{5}{7}-(-\frac{1}{7})=\frac{6}{7}$

Step-by-step explanation:

To evaluate :

$\frac{5}{7}-(-\frac{1}{7})$

Solution:

Two negatives multiply to become a positive.

Thus, we can remove parenthesis by reversing the signs of the fraction by multiplying the negative outside.

⇒ $\frac{5}{7}+\frac{1}{7}$

Since the denominators are same for both fractions, so we simply add the numerators.

⇒ $\frac{5+1}{7}$

⇒ $\frac{6}{7}$ (Answer)

6 0

3 years ago

The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circl

MAXImum [283]

Answer:

10 square inches per second.

Step-by-step explanation:

The radius of the circle is given by the equation:

r(t) = (1/π in/s)*t

Where time in seconds.

Remember that the area of a circle of radius R is written as:

A = π*R^2

Then the area of our circle will be:

A(t) = π*( (1/π in/s)*t)^2 = π*(1/π in/s)^2*(t)^2

Now we want to find the rate of change (the first derivation of the area) when the radius is equal to 5 inches.

Then the first thing we need to do is find the value of t such that the radius is equal to 5 inches.

r(t) = 5 in = (1/ in/s)*t

5in*(π s/in) = t

5*π s = t

So the radius will be equal to 5 inches after 5*π seconds, let's remember that.

Now let's find the first derivate of A(t)

dA(t)/dt = A'(t) = 2*(π*(1/π in/s)^2*t = (2*π*t)*(1/π in/s)^2

Now we need to evaluate this in the time such that the radius is equal to 5 inches, we will get:

A'(5*π s) = (2*π*5*π s)*((1/π in/s)^2

= (10*π^2 s)*(1/π^2 in^2/s^2) = 10 in^2/s

The rate of change is 10 square inches per second.

4 0

3 years ago

What line is parallel to 2x+5y=6 that goes through (5,3)

Anna11 [10]

Answer:

4y = x - 17

Step-by-step explanation:

First off, in the equation given : 2x + 5y = 6 , make y the subject of the formula which will give y = $\frac{-2x}{5}$ + $\frac{6}{5}$ which make the gradient, m, the coefficient of x = $\frac{-2}{5}$ . Since the line we're looking for is parallel to the one we were given, their respective gradients would be the same. (If they were perpendicular, the gradient of the new line would be a negative inverse of the given line)