2*8976543210= 17953086420
Answer:How do I round up one number how do I round up one number you word where not spell properly for the question
Step-by-step explanation:Rounding means making a number simpler but keeping its value close to what it was.
The result is less accurate, but easier to use.
round 73 is 70, 76 is 80
Example: 73 rounded to the nearest ten is 70, because 73 is closer to 70 than to 80. But 76 goes up to 80.
Common Method
Answer:
1;27
Step-by-step explanation:
I attached a photo of the work below :-)
Area of a trapezoid= [(a+b)÷2]h
1,224 in²= [(65.5 in+70.5in)÷2]h
1,224in²= (136÷2)h
1,224 in²= 68h
(1,224 in²÷68)= 68h÷68 (elimination)
h= 18 inches
Hi there!
![\large\boxed{(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty) }](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%28-%5Cinfty%2C%20%5Csqrt%5B3%5D%7B-4%7D%29%20%5Ctext%7B%20and%20%7D%20%280%2C%20%5Cinfty%29%20%7D)
We can find the values of x for which f(x) is decreasing by finding the derivative of f(x):

Taking the derivative gets:

Find the values for which f'(x) < 0 (less than 0, so f(x) decreasing):
0 = -8/x³ - 2
2 = -8/x³
2x³ = -8
x³ = -4
![x = \sqrt[3]{-4}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B-4%7D)
Another critical point is also where the graph has an asymptote (undefined), so at x = 0.
Plug in points into the equation for f'(x) on both sides of each x value to find the intervals for which the graph is less than 0:
f'(1) = -8/1 - 2 = -10 < 0
f'(-1) = -8/(-1) - 2 = 6 > 0
f'(-2) = -8/-8 - 2 = -1 < 0
Thus, the values of x are:
![(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C%20%5Csqrt%5B3%5D%7B-4%7D%29%20%5Ctext%7B%20and%20%7D%20%280%2C%20%5Cinfty%29)