Answer: There are 500 ninth grade students at Richbartville high school.
Step-by-step explanation:
Let be "x" the number of ninth grade students that are at Richbartville high school.
You know that the number of ninth grade students take algebra is 450 and this amount represents the 90% of the the ninth grade students at Richbartville high school.
Knowing this, you can set up the following proportion:
![\frac{90}{450}=\frac{100}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B90%7D%7B450%7D%3D%5Cfrac%7B100%7D%7Bx%7D)
The final step is to solve for "x" in order to find its value. This is:
![(x)(\frac{90}{450})=100\\\\x=(100)(\frac{450}{90})\\\\x=500](https://tex.z-dn.net/?f=%28x%29%28%5Cfrac%7B90%7D%7B450%7D%29%3D100%5C%5C%5C%5Cx%3D%28100%29%28%5Cfrac%7B450%7D%7B90%7D%29%5C%5C%5C%5Cx%3D500)
Therefore, there are 500 ninth grade students at Richbartville high school.
Answer:
65.200
Step-by-step explanation:
because if the number in the hundredths places is lower than 5 than the number in the tenth place stays the same and all of the numbers after the tenths place become a zero. hope i could help
Y = -x/5 I believe.
The negative reflects over the y-axis
and the coefficient 1/5 stretches
![y= ax^{2} -8x-3](https://tex.z-dn.net/?f=y%3D%20ax%5E%7B2%7D%20-8x-3)
1.
the line of symmetry is x=2, means that the x coordinate of the vertex is x=2.
the point x=2 is the midpoint of the roots
![x_1](https://tex.z-dn.net/?f=x_1)
and
![x_2](https://tex.z-dn.net/?f=x_2)
.
so
![\frac{x_1+x_2}{2}=2 ](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%3D2%0A%20)
![x_1+x_2=4](https://tex.z-dn.net/?f=x_1%2Bx_2%3D4)
Remark: in the x-axis, if c is the midpoint of a and b, then
![c= \frac{a+b}{2}](https://tex.z-dn.net/?f=c%3D%20%5Cfrac%7Ba%2Bb%7D%7B2%7D%20)
2.
since
![x_1](https://tex.z-dn.net/?f=x_1)
and
![x_2](https://tex.z-dn.net/?f=x_2)
are roots
![a(x_1)^{2} -8(x_1)-3=0](https://tex.z-dn.net/?f=a%28x_1%29%5E%7B2%7D%20-8%28x_1%29-3%3D0)
and
![a(x_2)^{2} -8(x_2)-3=0](https://tex.z-dn.net/?f=a%28x_2%29%5E%7B2%7D%20-8%28x_2%29-3%3D0)
3.
equalizing:
![a(x_1)^{2} -8(x_1)-3=a(x_2)^{2} -8(x_2)-3](https://tex.z-dn.net/?f=a%28x_1%29%5E%7B2%7D%20-8%28x_1%29-3%3Da%28x_2%29%5E%7B2%7D%20-8%28x_2%29-3)
![a(x_1)^{2} -8(x_1)=a(x_2)^{2} -8(x_2)](https://tex.z-dn.net/?f=a%28x_1%29%5E%7B2%7D%20-8%28x_1%29%3Da%28x_2%29%5E%7B2%7D%20-8%28x_2%29)
![a(x_1)^{2}-a(x_2)^{2} =8(x_1) -8(x_2)](https://tex.z-dn.net/?f=a%28x_1%29%5E%7B2%7D-a%28x_2%29%5E%7B2%7D%20%3D8%28x_1%29%20-8%28x_2%29)
in the left side factorize a, in the left side factorize 8:
![a[(x_1)^{2}-(x_2)^{2}] =8(x_1 -x_2)](https://tex.z-dn.net/?f=a%5B%28x_1%29%5E%7B2%7D-%28x_2%29%5E%7B2%7D%5D%20%3D8%28x_1%20-x_2%29)
in the right side use the difference of squares formula:
![a(x_1 -x_2)(x_1 +x_2) =8(x_1 -x_2)](https://tex.z-dn.net/?f=a%28x_1%20-x_2%29%28x_1%20%2Bx_2%29%20%3D8%28x_1%20-x_2%29)
simplify by
![(x_1 -x_2)](https://tex.z-dn.net/?f=%28x_1%20-x_2%29)
![a(x_1 +x_2) =8](https://tex.z-dn.net/?f=a%28x_1%20%2Bx_2%29%20%3D8)
substitute
![(x_1 +x_2)](https://tex.z-dn.net/?f=%28x_1%20%2Bx_2%29)
with 4:
![a*4 =8](https://tex.z-dn.net/?f=a%2A4%20%3D8)
a=2
Answer: C)2
The correct answer of the given question above would be 8ft/sec. The minimum speed that the stone be thrown so as to reach a height of 49 feet is 8ft/sec. Consider this given solution here:
u^2 = 2x a x 1
<span>u^2 = 2 x 32 x 1 </span>
<span>u = 8 ft/sec
</span>Hope this answer helps and this is what you are looking for.