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valina [46]
3 years ago
8

\times \frac{4}{7} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
torisob [31]3 years ago
5 0
Hi,
You simply need to multiply the numerator (4 in this case) with 5. Don’t touch the denominator (7)!
So the answer to this is 20 on 7 (20/7) or 2.857...
Butoxors [25]3 years ago
4 0

\bf 5\times \cfrac{4}{7}\implies \cfrac{5\cdot 4}{7}\implies \cfrac{20}{7}\implies \cfrac{14+6}{7}\implies \cfrac{14}{7}+\cfrac{6}{7}\implies 2+\cfrac{6}{7}\implies 2\frac{6}{7}

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In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC a
nirvana33 [79]

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ};

\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ};

\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q.

Also,

\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P in right triangle PBM.

Thus \rm \angle{P\hat{B}M} = \angle C\hat{M}Q.

Additionally \rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}.

Therefore \rm \triangle PBM \sim \triangle MCQ.

\rm \displaystyle BC = 2\;MC for M is the midpoint of segment BC.

\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC.

\rm \triangle PBM \sim \triangle MCQ implies that

\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3.

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