Create equations using the given information
1) x+y=56
2) y=x-16
Substitute the value of y into the 1st equation and simplify
1)x+x-16=56
2x=72
x=36
Substitute the value of x into the 1st equation and simplify
1)x+y=56
36+y=56
y=20
Answer: 20 and 36
51/100 would be larger
From the first look, I already know half of 100 is fifty and 51 is greater than that
So when I looked at 7/16, I was looking for something greater than half, but 7 is greater than half of 16.
51/100 is your answer
X + 8 >= 14
x >= 14 -8
x> = 6
Answer:
the length of the first side of the triangle is ![\frac{28}{3}\ in](https://tex.z-dn.net/?f=%5Cfrac%7B28%7D%7B3%7D%5C%20in)
the length of the second side of the triangle is ![\frac{37}{3}\ in](https://tex.z-dn.net/?f=%5Cfrac%7B37%7D%7B3%7D%5C%20in)
the length of the third side of the triangle is
Step-by-step explanation:
Let
x-----> the length of the first side of a triangle
y----> the length of the second side of a triangle
z---> the length of the third side of a triangle
we know that
-----> equation A
![x=z+3](https://tex.z-dn.net/?f=x%3Dz%2B3)
-----> equation B
The perimeter of the triangle is equal to
![P=x+y+z](https://tex.z-dn.net/?f=P%3Dx%2By%2Bz)
![P=28\ in](https://tex.z-dn.net/?f=P%3D28%5C%20in)
so
-----> equation C
substitute equation A and equation B in equation C
![28=x+(x+3)+(x-3)](https://tex.z-dn.net/?f=28%3Dx%2B%28x%2B3%29%2B%28x-3%29)
solve for x
![28=3x](https://tex.z-dn.net/?f=28%3D3x)
![x=\frac{28}{3}\ in](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B28%7D%7B3%7D%5C%20in)
Find the value of each side
the first side of a triangle is x
![x=\frac{28}{3}\ in](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B28%7D%7B3%7D%5C%20in)
the second side of a triangle is y
![y=\frac{28}{3}+3=\frac{37}{3}\ in](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B28%7D%7B3%7D%2B3%3D%5Cfrac%7B37%7D%7B3%7D%5C%20in)
the third side of a triangle is z