Answer:
Marco will need
of material to make the kite
Step-by-step explanation:
we know that
To know how much material Marco will need to make the kite, the area must be calculated.
Remember that the area of the kite is equal to
![A=\frac{1}{2}[d1*d2]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5Bd1%2Ad2%5D)
where
d1 and d2 are the diagonals of the kite
we have


substitute
![A=\frac{1}{2}[2*3]=3\ ft^{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5B2%2A3%5D%3D3%5C%20ft%5E%7B2%7D)
Answer:


Step-by-step explanation:
Given

Required
Solve:

Open bracket

Take LCM



Take LCM


Divide by 3/3

Answer:
I think the answer is about 10
Step-by-step explanation:
Answer:
13
Step-by-step explanation:
Write an equation setting the lengths equal to each other.
5x + 3 = 2x + 9
Move the variable (x) to one side. I'm going to subtract 2x from both sides.
5x - 2x + 3 = 2x - 2x + 9
3x + 3 = 9
Subtract 3 from both sides
3x +3 - 3 = 9 - 3
3x = 6
Divide both sides by 3
3x/3 = 6/3
x = 2
Now use 2x + 9 to find the length of EG by substituting 2 in for x.
2x + 9
2(2) + 9
4 + 9
13
You could also use 5x + 3 to find the length of EG by substituting 2 in for x.