Given:
Susan divides the fraction
by
.
Her friend Robyn divides
by
.
To find:
The quotient of Susan and Robyn.
Solution:
Susan divides the fraction
by
.
![\dfrac{\dfrac{5}{8}}{\dfrac{1}{16}}=\dfrac{5}{8}\times \dfrac{16}{1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdfrac%7B5%7D%7B8%7D%7D%7B%5Cdfrac%7B1%7D%7B16%7D%7D%3D%5Cdfrac%7B5%7D%7B8%7D%5Ctimes%20%5Cdfrac%7B16%7D%7B1%7D)
![\dfrac{\dfrac{5}{8}}{\dfrac{1}{16}}=\dfrac{80}{8}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdfrac%7B5%7D%7B8%7D%7D%7B%5Cdfrac%7B1%7D%7B16%7D%7D%3D%5Cdfrac%7B80%7D%7B8%7D)
![\dfrac{\dfrac{5}{8}}{\dfrac{1}{16}}=10](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdfrac%7B5%7D%7B8%7D%7D%7B%5Cdfrac%7B1%7D%7B16%7D%7D%3D10)
So, Susan's quotient is 10.
Her friend Robyn divides
by
.
![\dfrac{\dfrac{5}{8}}{\dfrac{1}{32}}=\dfrac{5}{8}\times \dfrac{32}{1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdfrac%7B5%7D%7B8%7D%7D%7B%5Cdfrac%7B1%7D%7B32%7D%7D%3D%5Cdfrac%7B5%7D%7B8%7D%5Ctimes%20%5Cdfrac%7B32%7D%7B1%7D)
![\dfrac{\dfrac{5}{8}}{\dfrac{1}{32}}=5\times 4](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdfrac%7B5%7D%7B8%7D%7D%7B%5Cdfrac%7B1%7D%7B32%7D%7D%3D5%5Ctimes%204)
![\dfrac{\dfrac{5}{8}}{\dfrac{1}{32}}=20](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdfrac%7B5%7D%7B8%7D%7D%7B%5Cdfrac%7B1%7D%7B32%7D%7D%3D20)
So, Robyn's quotient is 20.
Since 20>10, therefore, Robyn will get greater quotient.
34/12=2.888.....
2.888.....×10, 000=28, 333.333
28,333.333...ft or 28,333ft
Answer:
(8/5 , -2/5)
Step-by-step explanation:
x-y=2
2x+3y=14
-y=-x+2
y=x-2
2x+3(x-2)=14
2x+3x-6=14
5x-6=14
5x=8
x=8/5
y=(8/5)-2
y= -2/5
<u>ANSWER </u>
and ![y=\frac{1}{4}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B4%7D)
<u>EXPLANATION</u>
Given;
![4x-12y=1](https://tex.z-dn.net/?f=4x-12y%3D1)
and
![6x+4y=4](https://tex.z-dn.net/?f=6x%2B4y%3D4)
The augmented matrix of the two linear equation is given by;
![\left[\begin{array}{ccc}4&-12|&-1\\6\:&\:\:\:4|&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26-12%7C%26-1%5C%5C6%5C%3A%26%5C%3A%5C%3A%5C%3A4%7C%264%5Cend%7Barray%7D%5Cright%5D)
We now perform row operations;
.
This gives us
![\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\6\:&\:\:\:4|&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-3%7C%26%5Cfrac%7B-1%7D%7B4%7D%5C%5C6%5C%3A%26%5C%3A%5C%3A%5C%3A4%7C%264%5Cend%7Barray%7D%5Cright%5D)
![R_2-6R_1 \rightarrow R_2](https://tex.z-dn.net/?f=R_2-6R_1%20%5Crightarrow%20R_2)
![\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\0\:&\:\:\:22|&\frac{11}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-3%7C%26%5Cfrac%7B-1%7D%7B4%7D%5C%5C0%5C%3A%26%5C%3A%5C%3A%5C%3A22%7C%26%5Cfrac%7B11%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
![\frac{1}{22}R _2 \rightarrow R_2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B22%7DR%20_2%20%5Crightarrow%20R_2)
![\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\0\:&\:\:\:1|&\frac{1}{4}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-3%7C%26%5Cfrac%7B-1%7D%7B4%7D%5C%5C0%5C%3A%26%5C%3A%5C%3A%5C%3A1%7C%26%5Cfrac%7B1%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D)
![R_1+3R_2 \rightarrow R_1](https://tex.z-dn.net/?f=R_1%2B3R_2%20%5Crightarrow%20R_1)
![\left[\begin{array}{ccc}1&0|&\frac{1}{2}\\0\:&\:\:\:1|&\frac{1}{4}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%7C%26%5Cfrac%7B1%7D%7B2%7D%5C%5C0%5C%3A%26%5C%3A%5C%3A%5C%3A1%7C%26%5Cfrac%7B1%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D)
Hence
and ![y=\frac{1}{4}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B4%7D)