Square the top and the bottom. To get 7/11.
Finding the upper and lower bounds for a definite integral without an equation is pretty hard because how can we find the upper and lower bounds of definite integral if there is no equation given. But I will teach you how to find the lower and upper bounds of a definite integral, when the equation is like this
So, i integrate this,

I know I have a minimum at x=3 because;
f(t )= t^2 − 6t + 11
f′(t) = 2
t−6 = 0
2(t−3) = 0
t = 3
f(5) = 4
f(1) = −4
Answer:
Addison can eliminate $45 from Food/Clothes and $41 from Entertainment.
Step-by-step explanation:
To solve this problem you must apply the proccedure shown below:
1. Yoy have that the <span>rectangular tank has the following dimensions:
Length=3 inches
Widht=4 inches
Height=5 inches
2. Then, its volume is:
V1=LxWxH
V1=3 inchesx4 inchesx5 inches
V=60 inches</span>³
3. The sides of the <span>solid metal cube are of 3 inches, therefore, its volume is:
V2=s</span>³
V2=(3 inches)³
V2=27 inches³
4. T<span>he volume, in cubic inches, of water that the tank can now hold is:
Vt=V1-V2
Vt=60 inches</span>³-27 inches³
Vt=33 inches³