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djverab [1.8K]
2 years ago
5

Help please? requestig help to a smart person?

Mathematics
1 answer:
andrew-mc [135]2 years ago
8 0

<h3>Let's start by finding a ratio of Y to X</h3>

\frac{150}{10}  =  \frac{300}{20}  =  \frac{450}{30}  =  \frac{600}{40}  = 15

for \: x = 15y

50x = 50(15) = 750.00

<h2>Option B :)</h2>

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If QS bisects PQT, SQT=(8x- 25),PQT=(9x+34), and SQR=112, find each measure.
Goryan [66]

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)°

m∠PQT = (9·x + 34)°

m∠SQR = 112°

We have;

m∠PQT = m∠SQT + m∠PQS (Angle addition postulate)

m∠SQT ≅ m∠PQS (Angles formed by angle bisector are congruent)

m∠SQT = m∠PQS  by Definition of congruency

m∠PQT = 2 × m∠SQT

Therefore;

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

Collecting like terms gives:

(34 + 50)° = 16·x - 9·x = 7·x

7·x = 84°

x = 84°/7 = 12°

x = 12°

m∠SQT = (8·x - 25)°

Therefore;

m∠SQT = (8 × 12 - 25)° = 71°

m∠SQT = 71°

m∠PQT = 2 × m∠SQT

∴ m∠PQT = 2 × 71° = 142°

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°

Learn more about angles here:

brainly.com/question/2882938

#SPJ1

4 0
2 years ago
50 points! I will mark brainliest! Random answers will be reported!
marta [7]

Answer:

The answer is c.

Step-by-step explanation:

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2 years ago
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