For one of the triangular sides, 2.6 x 4 = 10.4 divided by 2 = 5.2
Other triangle; 2.6 x 3 = 7.8 /2 = 3.9
Rectangular side in front; 6.5 x 3 = 19.5
Back; 6.5 x 4 = 26
Base; 6.5 x 2.6 = 16.9
Answer: For the surface area it is 74.6 (SUPER SORRY IF THIS ISNT RIGHT OMG)
Answer:
F. 8
Step-by-step explanation:
The ratio of the long side to the short side is the same in similar triangles. The long side of triangle BAD is AD, which has length 20-4 = 16.
BD/DE = AD/BD
h/4 = 16/h
h^2 = 64 . . . . . . . multiply by 4h
h = 8 . . . . . . . . . . take the square root (matches selection F)
_____
<em>Comment on this geometry</em>
BD = √(AD·DC) is called the "geometric mean" of the segments AD and DC. This geometry has some other geometric mean relationships as well:
BC = √(AC·DC)
BA = √(AC·AD)
Answer:
2 pounds.
Step-by-step explanation:
We have been given that Luke bought 256 ounces of strawberries. He divided them evenly between 8 different pies. We are asked to find the pounds of strawberries that Luke put on each pie.
First of all, we will convert 256 ounces into pounds.
We know that 1 pound equals 16 ounces. To convert 256 ounces into pounds, we need to divide 256 by 16.

Now we will divide 16 pounds by 8 to find number of strawberries put on each pie.


Therefore, Luke put 2 pounds strawberries on each pie.
Answer:
Step-by-step explanation:
Given
The sum of the two positive integer a and b is at least 30, this means the sum of the two positive integer is 30 or greater than 30, so we write the inequalities as below.
The difference of the two integers is at least 10, if b is the greater integer then we subtract integer a from integer b, so we write the inequality as below.
Therefore, the following system of inequalities could represent the values of two positive integers a and b.
we'll start off by grouping some

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)