It would pend on whetch one because you devide some nummernummer tell me witch
14 / 21 = X/15 => X = 14 * 15 / 21 => divide by 7 => 2 * 15 / 3 => divide by 3 => 2 * 5 / 1 => 10
Hi there!
![\large\boxed{53) \text{ } KL = 31}\\\\\large\boxed{54) \text{ } { } DC = 3}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B53%29%20%5Ctext%7B%20%7D%20KL%20%3D%2031%7D%5C%5C%5C%5C%5Clarge%5Cboxed%7B54%29%20%5Ctext%7B%20%7D%20%7B%20%7D%20DC%20%3D%203%7D)
53)
The altitude of a triangle is always perpendicular to the base, therefore:
5x° must equal 90°.
Solve for x:
5x = 90
5x / 5 = 90 / 5 ----> x = 18
KL is equal to 2x - 5, so plug the value of x into the equation:
KL = 2(18) - 5
KL = 36 - 5 ---> KL = 31
54)
A median bisects the base, therefore:
AD ≅ DC
Set both equations equal to each other:
3x = 2x + 1
Subtract 2x from both sides:
x = 1
Plug in the value of x into the equation for DC:
2(1) + 1 = 3.
DC = 3.
The largest square you can make using 40 tiles, would be 100
Answer:
![2x^2-12x+22](https://tex.z-dn.net/?f=2x%5E2-12x%2B22)
Step-by-step explanation:
Let's follow what the problem asks us:
- "The area of a square with side length x"
this is the formula for area: ![area=side*side=side^2](https://tex.z-dn.net/?f=area%3Dside%2Aside%3Dside%5E2)
since the side length is x, the initial area is: ![x^2](https://tex.z-dn.net/?f=x%5E2)
- Now, "where the side length is decreased by 3":
The side is now
, thus the area is: ![(x-3)^2=x^2-6x+9](https://tex.z-dn.net/?f=%28x-3%29%5E2%3Dx%5E2-6x%2B9)
- "the area is multiplied by 2"
we had the area
, when we multiply by 2:
![(2)(x^2-6x+9)=2x^2-12x+18](https://tex.z-dn.net/?f=%282%29%28x%5E2-6x%2B9%29%3D2x%5E2-12x%2B18)
- And finally " then 4 square units are added to the area."
this is: ![2x^2-12x+18 + 4=2x^2-12x+22](https://tex.z-dn.net/?f=2x%5E2-12x%2B18%20%2B%204%3D2x%5E2-12x%2B22)