Base of prism is seemed to be a triangle
- Base=B=3
- Height of base=H=6units
- Height of prism=8
Area of base=


Now

Answer:
-2 11/35
Step-by-step explanation:
Hope it helps :)
Hello!
To solve an algebraic equation, you must isolate the variable on one side. We’ll begin by restating the given equation:
12 + 12n = -48
Now subtract 12 from both sides of the equation:
12n = -60
Divide both sides by 12:
n = -5
We have now proven that “n” is equal to (-5).
I hope this helps!
Answer: x = 7 or x = -8
x² + x - 56 = 0
⇔ x² + 8x - 7x - 56 = 0
⇔ x(x + 8) - 7(x + 8) = 0
⇔ (x - 7)(x + 8) = 0
⇔ x - 7 = 0
or x + 8 = 0
⇔ x = 7 or x = -8
Step-by-step explanation:
1.) The interval of the value of x is from -5 to 1, inclusive. Remember that what is asked is the absolute value, thus the sign does not matter even if you have to subtract x from 5. Thus, the maximum value would be obtained if the x is smaller, which is 1. The minimum value is obtained when x=-5.
Absolute maximum value:
x = - 5f(-5) = ║5 - 7(-5)^2║ = ║-170║=
170Absolute minimum value:
x = 1f(1) = ║5 - 7(1)^2║ = ║-2║=
2
2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:

The differentiated form would be f'(x) = -2x. Then,


Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.