Yes because say we are doing 2/10 if you divide 2 by 2 and 10 by 2 it will be 1/5 which equals the first one so yes they are.
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Answer:
12/5 as an improper fraction
2 2/5 as a mixed number
Step-by-step explanation:
3 * 4/5
Rewriting the whole number as a fraction
3/1 * 4/5
Multiply the numerators
3*4 = 12
Multiply the denominators
1*5 =5
Numerator over denominators
12/5
Rewriting as a mixed number
5 goes into 12 2 times (2*5=10 12-10=2) with 2 left over
2 2/5
let's firstly convert the mixed fraction to improper fraction and then multiply.
![\bf \stackrel{mixed}{3\frac{2}{5}}\implies \cfrac{3\cdot 5+2}{5}\implies \stackrel{improper}{\cfrac{17}{5}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{17}{~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\implies 17](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B2%7D%7B5%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%205%2B2%7D%7B5%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B17%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B17%7D%7B~~%5Cbegin%7Bmatrix%7D%205%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%5Ccdot%20~~%5Cbegin%7Bmatrix%7D%205%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%5Cimplies%2017)
Answer:
6(p+4q)
Step-by-step explanation:
