the answer is the first one A
Answer:convergent
Step-by-step explanation:
Given
Improper Integral I is given as
integration of 
is
![I=1000\times \left [ e^x\right ]^{0}_{-\infty}](https://tex.z-dn.net/?f=I%3D1000%5Ctimes%20%5Cleft%20%5B%20e%5Ex%5Cright%20%5D%5E%7B0%7D_%7B-%5Cinfty%7D)
![I=1000\times I=\left [ e^0-e^{-\infty}\right ]](https://tex.z-dn.net/?f=I%3D1000%5Ctimes%20I%3D%5Cleft%20%5B%20e%5E0-e%5E%7B-%5Cinfty%7D%5Cright%20%5D)
![I=1000\times \left [ e^0-\frac{1}{e^{\infty}}\right ]](https://tex.z-dn.net/?f=I%3D1000%5Ctimes%20%5Cleft%20%5B%20e%5E0-%5Cfrac%7B1%7D%7Be%5E%7B%5Cinfty%7D%7D%5Cright%20%5D)
so the integration converges to 1000 units
Answer: 7
Step-by-step explanation:
They are alternate interior angle
2x + 5 = 3x -2
2x = 3x -7
-x = -7
x = 7
We know that the sandbox is square, and that the area is 160 sq ft. Since A = s^2 for a square, let's work backwards and find s if A = 160 sq ft = s^2.
s^2 = 160 = 16(10). We do this because 16 is the largest perfect square factor of 160.
Then the length of one side of the square is approx. s = 4√10 ft (12.65 ft)
Answer:
Option D. (x + 4)(x + 1)
Step-by-step explanation:
From the question given above, the following data were obtained:
C = (6x + 2) L
D = (3x² + 6x + 9) L
Also, we were told that half of container C is full and one third of container D is full. Thus the volume of liquid in each container can be obtained as follow:
Volume in C = ½C
Volume in C = ½(6x + 2)
Volume in C = (3x + 1) L
Volume in D = ⅓D
Volume in D = ⅓(3x² + 6x + 9)
Volume in D = (x² + 2x + 3) L
Finally, we shall determine the total volume of liquid in the two containers. This can be obtained as follow:
Volume in C = (3x + 1) L
Volume in D = (x² + 2x + 3) L
Total volume =?
Total volume = Volume in C + Volume in D
Total volume = (3x + 1) + (x² + 2x + 3)
= 3x + 1 + x² + 2x + 3
= x² + 5x + 4
Factorise
x² + 5x + 4
x² + x + 4x + 4
x(x + 1) + 4(x + 1)
(x + 4)(x + 1)
Thus, the total volume of liquid in the two containers is (x + 4)(x + 1) L.
