Find the values of b such that the function has the given minimum value. f(x) = x2 + bx − 21;
Minimum value:
−70

b = 14.

Step-by-step explanation:

We have the function:

f(x) = x^2 + b*x - 21

The minimum of a quadratic function will be at the vertex, and the vertex is at the value of x such that:

f'(x) = 2*x + b = 0

then the vertex is at:

x = -b/2

Then the minimum of the function f(x) will be:

f(-b/2) = (-b/2)^2 + b*(-b/2) - 21

and we know that the minimum value is -70, then:

f(-b/2) = -70 = (b^2)/4 - (b^2)/2 - 21

Now we need to solve this equation for b.

-70 = (b^2)/4 - (b^2)/2 - 21

-70 = -(b^2)/4 - 21

-70 + 21 = -(b^2)/4

49 = (b^2)/4

49*4 = b^2

196 = b^2

√196 = b = 14

The value of b is 14.

ddd

step-by-step explanation:

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240