Interesting that only integrals along the
-axis are suggested when integrating along the
-axis would be much simpler... Anyway, you have to split the interval of integration into two. The "height" of the region is not uniform over the entire interval.
When
, we have
. When
, we have
. Then the area we want is given by
![\displaystyle\int_0^12\,\mathrm dx+\int_1^52-\sqrt{x-1}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E12%5C%2C%5Cmathrm%20dx%2B%5Cint_1%5E52-%5Csqrt%7Bx-1%7D%5C%2C%5Cmathrm%20dx)
which seems to agree with the last option.
Answer:
negative correlation on Edg
Step-by-step explanation:
65x+75=495
-75 -75
65x=420
65/65 420/65
X=6.46
Answer:
4
3
+
5
9
2
−
8
0
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
Area of a triangle:
![A=\frac{b*h}{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7Bb%2Ah%7D%7B2%7D)
In our case:
b=4
h=2
Plug in what we know:
![A=\frac{(4)(2)}{2} \\A=4units](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%284%29%282%29%7D%7B2%7D%20%5C%5CA%3D4units)
Find the matching solution:
A.) it is 1/2 the area of a rectangle of length 4 units and width 2 units
X B.) it is twice the area of a rectangle of length 4 units and width two units
X C.) it is 1/2 the area of a square of side length 4 units
X D.) it is twice the area of a square of side length 4 units