First find the length of the diagonal of a square of side 4 inches:
d^2 = 4^2 + 4^2 = 2*4^2 = 32. Then the diagonal of the cube has length
sqrt( 32 + 4^2) = sqrt(32+16) = sqrt(48) = 4sqrt(3).
1. N + 2; n = 6: the answer is 8 because n=6 so we sub n with that... and we get 6+2 and that’s 8.
2.5f; where f=4: the answer is 20 because when a variable is directly next to a number it is multiplied by that number so we will replace f with 4 and our equation is now 5(4) or 5•4 and both are equivalent to 20.
3. 7b-2; where b=5: the answer is 33 because 7 multiplied by 5(b) minus 2= 7(5)-2 or 7•5-2= 33 because you will multiply 7 by 5 and get 33 then you will subtract by 2 and get 33.
Hope this helps!!!
Answer:
Whats the question
Step-by-step explanation:
Answe
Given,
f(x) = 49 − x² from x = 1 to x = 7
n = 4
For x= 1
f(x₀) = 49 - 1^2 = 48
x = 2.5
f(x₁) = 42.75
x = 4
f(x₂) = 49 - 4^2 = 33
x = 5.5
f(x₃) = 49 - 5.5^2 = 18.75
x = 7
f(x₄) = 49 - 7^2 = 0
We have to evaluate the function on therigh hand point
![A = \Delta x [f(x_1)+f(x_2)+f(x_3)+f(x_4)]](https://tex.z-dn.net/?f=A%20%3D%20%5CDelta%20x%20%5Bf%28x_1%29%2Bf%28x_2%29%2Bf%28x_3%29%2Bf%28x_4%29%5D)
![A = 1.5 [42.75+33+18.75+0]](https://tex.z-dn.net/?f=A%20%3D%201.5%20%5B42.75%2B33%2B18.75%2B0%5D)
For Area for left hand sum
![A = \Delta x [f(x_0)+f(x_1)+f(x_2)+f(x_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5CDelta%20x%20%5Bf%28x_0%29%2Bf%28x_1%29%2Bf%28x_2%29%2Bf%28x_3%29%5D)
![A = 1.5 [48+42.75+33+18.75]](https://tex.z-dn.net/?f=A%20%3D%201.5%20%5B48%2B42.75%2B33%2B18.75%5D)
