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Novosadov [1.4K]
4 years ago
14

Determine which expressions is an equivalent form of c/13a select all situations that apply

Mathematics
1 answer:
kompoz [17]4 years ago
7 0

Answer:

  \dfrac{5a^3bc^2}{65a^4bc}

Step-by-step explanation:

Of the given choices A, D, E have the correct constant. D has an exponent of <em>a</em> that is too large. E has <em>a</em> in the numerator, not the denominator. So, the only viable choice is the first answer selection.

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x² - 3x + 30 = 8x         subtract 8x from both sides ⇒<span>
    - 8x          - 8x
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x</span>² - 11x + 30 = 0    ←  <span>standard form
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A. Add 3 to both sides and subtract 8x from both sides.
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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

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4 0
2 years ago
Find the complex zeros of <br><img src="https://tex.z-dn.net/?f=%20f%28x%29%20%3D%20%7Bx%7D%5E%7B3%20%7D%20%20-%2013%20%7Bx%7D%5
timama [110]

f(x) = 0 \\\\\implies x^3-13x^2+59x -87 =0\\\\\implies x^3 -3x^2 -10x^2 +30x +29x - 87=0\\\\\implies x^2(x-3)  - 10x(x-3) + 29(x-3) =0\\\\\implies (x-3)(x^2 -10x +29) =0\\\\\implies x - 3 = 0 ~~ \text{or}~~x^2 -10x +29 = 0\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{-(-10) \pm \sqrt{(-10)^2 -4 \cdot 1 \cdot 29}}{2(1)}\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{10\pm \sqrt{-16}}{2}\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{10\pm 4i }{2}\\

\\\implies x =3~~ \text{ or} ~~ x =5 \pm 2i\\\\\text{Hence the complex roots are,}~ 5- 2i ~~ \text{and} ~~5 + 2i

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