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Darya [45]
3 years ago
12

Jess is comparing fractions, which fraction is greater than 5/6? A) 7/8B) 4/5C) 3/4D) 2/3

Mathematics
2 answers:
Yuki888 [10]3 years ago
8 0

Answer:

7/8 > 5/6

Step-by-step explanation:

A) 7/8

We can compare this as follows.

Lets say both are equal. \frac{5}{6}= \frac{7}{8}

Cross multiplying these we get 40=42

We get 40<42. In fraction we get \frac{5}{6}

In case if you want to convert this to decimal, we get;

5/6 = 0.833 and 7/8 = 0.875

We get 5/6<7/8

B) 4/5

Similarly we get 4/5 = 0.8 and 5/6 = 0.833

Here 4/5<5/6

C) 3/4

we get 3/4 = 0.75 and 5/6 = 0.833

3/4<5/6

D) 2/3

we get 2/3 = 0.66 and 5/6 = 0.833

2/3<5/6

levacccp [35]3 years ago
6 0
I think the answer is D

                                                                    
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Step-by-step explanation:

2x - y = -7

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Answer:

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Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

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\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

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=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

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