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vladimir2022 [97]
3 years ago
9

How Many 4/3 are in 8

Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
6 0
8 divide by 4/3

To divide<span> by a </span>factor<span>, </span>multiply<span> by its </span>reciprocal<span>.
</span><span>8(<span>3/4</span>)
</span>
<span>Cancel 4 and 8.
</span><span>2⋅3
</span>
Multiply 2<span> by </span>3<span> to get </span><span>6.
</span>6

There are "6" 4/3's in 8
professor190 [17]3 years ago
5 0
= 8 / 4/3 
= 8 * 3/4
= 6

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If QS bisects PQT, SQT=(8x- 25),PQT=(9x+34), and SQR=112, find each measure.
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The values of the angles are; x = 12°, m∠PQS = 71°, m∠PQT = 142° and m∠TQR = 41°

<h3>What are the measure of the each angle?</h3>

The required angles; x, m∠PQS, m∠PQT, and m∠TQR

The given parameters are bisects ∠PQT

m∠SQT = (8·x - 25)°

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m∠SQR = 112°

We have;

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(9·x + 34)° = 2 × (8·x - 25)° = (16·x - 50)°

Collecting like terms gives:

(34 + 50)° = 16·x - 9·x = 7·x

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x = 12°

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Therefore;

m∠SQT = (8 × 12 - 25)° = 71°

m∠SQT = 71°

m∠PQT = 2 × m∠SQT

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m∠PQT = 142°

m∠PQS = m∠SQT (Angles formed by the same bisector )

∴ m∠PQS = m∠SQT = 71°

m∠PQS = 71°

m∠SQR = m∠SQT + m∠TQR (Angle addition postulate)

m∠SQT = 71°

∴  m∠SQR = 112° = 71° + m∠TQR

m∠TQR = 112° - 71° = 41°

m∠TQR = 41°

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brainly.com/question/2882938

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