True for #1. False for he rest of them.
Check the picture below.
so the surface area of the pyramid will be the sum of the areas of the base and the four triangular faces.
![\stackrel{\textit{\Large Areas}}{\stackrel{\textit{4 triangular faces}}{4\left[ \cfrac{1}{2}(7.5)(16) \right]}~~ + ~~\stackrel{\textit{rectangular base}}{(7.5)(7.5)}}\implies 240~~ + ~~56.25\implies 296.25~ft^2](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7B%5Ctextit%7B4%20triangular%20faces%7D%7D%7B4%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%287.5%29%2816%29%20%5Cright%5D%7D~~%20%2B%20~~%5Cstackrel%7B%5Ctextit%7Brectangular%20base%7D%7D%7B%287.5%29%287.5%29%7D%7D%5Cimplies%20240~~%20%2B%20~~56.25%5Cimplies%20296.25~ft%5E2)
12.
It is easy all you have to do is 15-3 because 1+2=3
Answer:
A and C.
Step-by-step explanation:
to get rid of the square root & n, you would have to multiply it by n on both sides of the equation, giving you r^n=a.
also to get rid of the square root & n, you can also multiply both sides of a and r by the exponential fraction with n as the denominator:
![\sqrt[n]{a} =r \\(a)^\frac{1}{n} = r\\a\frac{1}{n} = r](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%3Dr%20%5C%5C%28a%29%5E%5Cfrac%7B1%7D%7Bn%7D%20%3D%20r%5C%5Ca%5Cfrac%7B1%7D%7Bn%7D%20%3D%20r)
Answer:
-1 and -2, or 9 and 10
Step-by-step explanation:
If x is the smaller integer:
x (x + 1) = 14 + 4 (x + x + 1)
x² + x = 14 + 4 (2x + 1)
x² + x = 14 + 8x + 4
x² − 7x − 18 = 0
Factor using AC method:
(x − 9) (x + 2) = 0
x = -2 or 9
x is the smaller integer, so the larger integers are -1 and 10. Check each pair of solutions.
-1 and -2:
(-2) (-1) = 14 + 4 (-2 + -1)
2 = 14 + 4 (-3)
2 = 2
9 and 10:
(9) (10) = 14 + 4 (9 + 10)
90 = 14 + 4 (19)
90 = 90
Both solutions work.