For this case we have that the quotient of 6 and a number, can be expressed as:
![\frac {6} {x}](https://tex.z-dn.net/?f=%5Cfrac%20%7B6%7D%20%7Bx%7D)
Where the variable "x" represents the incognito number.
Now we have that expression is subtracted from 100. Now, we can write the following:
![100- \frac {6} {x}](https://tex.z-dn.net/?f=100-%20%5Cfrac%20%7B6%7D%20%7Bx%7D)
ANswer:
Option A
0.000000000093 in scientific notation is 9.3 x 10 ^ -11
Answer:
105 maybe
Step-by-step explanation:
Answer: ![D'(-14,1);\ E'(4,7);\ F'(4,1)](https://tex.z-dn.net/?f=D%27%28-14%2C1%29%3B%5C%20E%27%284%2C7%29%3B%5C%20F%27%284%2C1%29)
Step-by-step explanation:
Since the center of dilation is not at the origin, we can use the following formula in order to find the coordinates of the vertices of the triangle D'E'F':
![D_{O,k}(x,y)=(k(x-a)+a, k(y-b)+b)](https://tex.z-dn.net/?f=D_%7BO%2Ck%7D%28x%2Cy%29%3D%28k%28x-a%29%2Ba%2C%20k%28y-b%29%2Bb%29)
Where "O" is the center of dilation at (a,b) and "k" is the scale factor.
In this case you can identify that:
![(a,b)=(1,1)\\k=3](https://tex.z-dn.net/?f=%28a%2Cb%29%3D%281%2C1%29%5C%5Ck%3D3)
Therefore, susbtituting values into the formula shown above, you get that the coordinates ot the resulting triangle D'E'F, are the following:
Vertex D' → ![(3(-4-1)+1,\ 3(1-1)+1)=(-14,1)](https://tex.z-dn.net/?f=%283%28-4-1%29%2B1%2C%5C%203%281-1%29%2B1%29%3D%28-14%2C1%29)
Vertex E' → ![(3(2-1)+1,\ 3(3-1)+1)=(4,7)](https://tex.z-dn.net/?f=%283%282-1%29%2B1%2C%5C%203%283-1%29%2B1%29%3D%284%2C7%29)
Vertex F' → ![(3(2-1)+1,\ 3(1-1)+1)=(4,1)](https://tex.z-dn.net/?f=%283%282-1%29%2B1%2C%5C%203%281-1%29%2B1%29%3D%284%2C1%29)
Nickels is 0,05 of dolar - n
dimes is 0,1 o dolar -d
quarters is 0,25 of dolar -q
dimes =2nickels
d=2n
quarters=2dimes
q=2d=4n
12,5=n(0,05)+2n(0,1)+4n(0,25)
12,5=0,05n+0,2n+n
12,5=1,25n
n=10
q=4n=40
There are 40 quarters