Answer:
<h3>5</h3>
Step-by-step explanation:
Given the expression
2a^3−10ab^2+3a^3−ab^2−7
We are to find the coefficient of a^3
First is to collect the like terms;
2a^3−10ab^2+3a^3−ab^2−7
= 2a^3+3a^3−10ab^2−ab^2−7
= 5a^3-11ab^2-7
From the resulting equation, you can see that the coefficient of the term having a^3 is 5
<span>6( 5-8x) +12= -54
30-48x+12=-54 this comes from distributing(multiplying) 6x5 and 6 times -8x
30+12-48x=-54 you have to add the coefficients 30+12
42-48x=-54
-42 -42
-----------------------
-48x=-96 divide by -48
x=2</span>
a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so

Then by linearity, we have
