Hi there!
We can evaluate using the power rule and trig rules:
Therefore:
![\int\limits^{12}_{2} {x-sin(x)} \, dx = [\frac{1}{2}x^{2}+cos(x)]_{2}^{12}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B12%7D_%7B2%7D%20%7Bx-sin%28x%29%7D%20%5C%2C%20dx%20%3D%20%5B%5Cfrac%7B1%7D%7B2%7Dx%5E%7B2%7D%2Bcos%28x%29%5D_%7B2%7D%5E%7B12%7D)
Evaluate:
Answer:
75
Step-by-step explanation:
Answer:Area of part 1:
A1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2
Area of part 2:
A2=length*width=(8+2)*1=10 ft2
Area of part 3:
A3=length*width=8*2=16 ft2
Total shaded area : A1+A2+A3=41 ft2
Total area of the reactangle : A=16*8=128 ft2
Total area of the nonshaded region : 128-41=87 ft2
Step-by-step explanation: Plz make brainliest
Answer:
Step-by-step explanation:
bisects 
and 
.
Solve for m.
Subtract 3m from both sides.
Divide both sides by 3.
Calculate 
.
Put m = 7.
Answer:
A perfect square is a whole number that is the square of another whole number.
n*n = N
where n and N are whole numbers.
Now, "a perfect square ends with the same two digits".
This can be really trivial.
For example, if we take the number 10, and we square it, we will have:
10*10 = 100
The last two digits of 100 are zeros, so it ends with the same two digits.
Now, if now we take:
100*100 = 10,000
10,000 is also a perfect square, and the two last digits are zeros again.
So we can see a pattern here, we can go forever with this:
1,000^2 = 1,000,000
10,000^2 = 100,000,000
etc...
So we can find infinite perfect squares that end with the same two digits.
