9514 1404 393
Answer:
b. 17°
Step-by-step explanation:
All of the angles are 2nd-quadrant angles except 17°, which is a 1st-quadrant angle.
17° is the odd angle
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When you compute ...
mod(angle, 360), you get 163, 17, 163, 163.
4t=r
a=pir^2
sub 4t for r
a=pi(4t)^2
a=pi16t^2
a(t)=16pi(t^2)
A. a(t)=16pi(t^2)
B. sub 4 for t
a(4)=16pi4^2
a(4)=16pi16
a(4)=16*16*3.14
a(4)=803.84 square units
A. a(t)=16pi(t^2)
B. 803.84 square units
We are given the area of the region under the curve of the function f(x) = 5x + 7 with an interval [1, b] which is 88 square units where b > 1
We need to find the integral of the function f(x) = 5x + 7 with the limits 1 and b
5/2 x^2 + 7x (limits: 1, b)
substitute the limits:
5/2 (1^2) + 7 (1) - 5/2 b^2 + 7b = 0
solve for b
Then after solving for b, this would be your interval input with 1: [1, b].<span />
12 to 16 because all ratios are equivalent
