The answer is 5. Work backwards: first add 7 to 9, then divide by 2, then subtract 3.

9+7=16
16/2=8
8-3=5

Cross Check:
5+3=8
8*2=16
16-7=9

Final Answer: 5

4 0

3 years ago

Write a definite integral that represents the area of the region. (Do not evaluate the integral.)

joja [24]

Answer:

$\int_{-1}^{1}7(x^{3}-x)\:dx$

Step-by-step explanation:

1) The other curve is $y=0$ then the common points of both curves are x-intercepts, the roots of $y=7(x^{3}-x)$

$y=7(x^{3}-x)\Rightarrow 7(x^3-x)=0 \Rightarrow 7(x^{3}-x)=7x(x-1)(x-1)\Rightarrow \\S=\left ( 0,0 \right ),\left ( -1,0 \right ),\left ( 1,0 \right )$

2). Then those intersection points are the upper and the lower limits. Plugging in to this formula for they belong to the interval [-1,1]:

$\int_{a}^{b}|f(x)-g(x)dx$

$\int_{a}^{b}|f(x)-g(x)dx \Rightarrow \int_{-1}^{1}|7(x^{3}-x)-0|dx \Rightarrow \int_{-1}^{1}7(x^{3}-x)\:dx$

8 0

3 years ago

Which of the point-slope equations below are correct for the line that passes through points (-2, -2) and (2, 1)?

Ne4ueva [31]

Find slope first:
m =y2-y1/x2-x1
m = 1--2/2--2
m = 3/4

Select a point, insert your slope, & put it into point slope form:
y-y1 =m(x-x1)

Your final answers are:

y-1=3/4(x-2)
OR
y+2 =3/4(x+2)

8 0

3 years ago