Find the GCF of 80 and 32.
I'd start by identifying possible integer factors of both 80 and 32:
80: {1,2,4,5,8,10,16,20, 40, 80}
32: {1, 2,4, 8, 16, 32}
Working backwards, we see that the first factor that is represented in both lists is 16. Is 80 evenly divisible by 16? Yes; the quotient is 5.
Is 32 evenly divisible by 16? Yes; the quotient is 2.
You could writet 80 + 32 as 16(5 + 2). This is a product equal to 112, just as 80 + 32 = 112.
If x=width, then the equation would be:
x(x+10)=20,000
The length is 10 meters longer than the width(x), so the length is x+10.
To get the area(20,000) you multiply width by length, hence x(x+10).
Hope this helps.
So,first step is to write ![(fog)(-4)) =f[g(-4)] \\](https://tex.z-dn.net/?f=%20%28fog%29%28-4%29%29%20%3Df%5Bg%28-4%29%5D%20%5C%5C%20)
Now we start from inner paranthesis
,we need to first find value of
![g(-4) =\frac{\sqrt{12-(-4)}}{3}\\ =\frac{\sqrt{12+4}}{3}\\ =\frac{\sqrt{16}}{3}\\ =\frac{4}{3}\\ (fog)(-4)) =f[g(-4)] =f(\frac{4}{3}) =9(4/3)^{2} =9*(16/9) =144/9 =16](https://tex.z-dn.net/?f=%20g%28-4%29%20%3D%5Cfrac%7B%5Csqrt%7B12-%28-4%29%7D%7D%7B3%7D%5C%5C%20%3D%5Cfrac%7B%5Csqrt%7B12%2B4%7D%7D%7B3%7D%5C%5C%0A%3D%5Cfrac%7B%5Csqrt%7B16%7D%7D%7B3%7D%5C%5C%0A%3D%5Cfrac%7B4%7D%7B3%7D%5C%5C%0A%28fog%29%28-4%29%29%20%3Df%5Bg%28-4%29%5D%20%3Df%28%5Cfrac%7B4%7D%7B3%7D%29%20%3D9%284%2F3%29%5E%7B2%7D%20%3D9%2A%2816%2F9%29%20%3D144%2F9%20%3D16%20)
Answer:
12.5 ft
Step-by-step explanation:
30 : x = 24 :10
30*10 = 24x
x= 300/24 =12.5
If you need the y zero, you would take the 2 points on the graph of (3, 0) and (6, 0) so the 2 zero’s are 3, and 6.
