The lowest term is 
.
Solution:
Given expression is 
<u>To reduce this term to the lowest term:</u>
Multiply the numerator and denominator.
Now, divide the numerator and denominator by the greatest common factor.
Here 150 and 8 both have common factor 2.
So, divide numerator and denominator by 2.
Hence the lowest term is .
<h3>
Answer: -20</h3>
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Work Shown:
Let x be the location of E on the number line.
Since C is the midpoint of E and F, this means we can find C's location by adding E and F together and dividing that sum by 2
midpoint = (endpoint1 + endpoint2)/2
C = (E+F)/2
Plug in E = x, C = -8 and F = 4. Then solve for x
C = (E+F)/2
-8 = (x+4)/2
(x+4)/2 = -8
x+4 = 2(-8) .... multiplying both sides by 2
x+4 = -16
x = -16-4 .... subtract 4 from both sides
x = -20
The location of point E on the number line is -20
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As a check, lets add E and F to get E+F = -20+4 = -16
Then cut this in half to get -16/2 = -8, which is the proper location of point C
This confirms our answer.
Answer:
( 2,1) is the center of dilation and -2 is the scale factor
Step-by-step explanation:
We can use the formula
A' = k( x-a) +a, k( y-b)+b where ( a,b) is the center of dilation and k is the scale factor
(0,0) becomes (6,3)
( 6,3) = k( 0-a) +a, k( 0-b)+b
6 = -ka+a
3 = -kb+b
We also have
(4,0) becomes (-2,3)
( -2,3) = k( 4-a) +a, k( 0-b)+b
-2 =4k -ka+a
3 = -kb+b
Using these two equations
6 = -ka+a
-2 =4k -ka+a
Subtracting the top from the bottom
-2 =4k -ka+a
-6 = ka -a
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-8 = 4k
Divide by 4
-8/4 = 4k/4
-2 = k
Now solving for a
6 = -ka +a
6 = - (-2)a +a
6 = 2a+a
6 = 3a
Divide by 3
6/3 =3a/3
2=a
Now finding b
3 = -kb+b
3 = -(-2)b+b
3 = 2b+b
3 = 3b
b=1
a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cqquad%20%5Cleftarrow%20vertical%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D0%5C%5C%20k%3D0%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-0%29%5E2%2B0%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20x%3D-2%5C%5C%20y%3D3%20%5Cend%7Bcases%7D%5Cimplies%203%3Da%28-2-0%29%5E2%2B0%5Cimplies%203%3D4a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B3%7D%7B4%7D%3Da~%5Chspace%7B10em%7Dy%3D%5Ccfrac%7B3%7D%7B4%7D%28x-0%29%5E2%2B0%5Cimplies%20%5Cboxed%7By%3D%5Ccfrac%7B3%7D%7B4%7Dx%5E2%7D)
Answer:
4.31 kilometer
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