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sweet [91]
4 years ago
10

∠rst and ∠vst are a linear pair, m∠rst = 3x 7, and m∠vst = 9x 17. what are the measures of m∠rst and m∠vst?a.46, 134b.26, 154c.5

.5, 84.5d.58, 122
Mathematics
1 answer:
sergij07 [2.7K]4 years ago
6 0
M∠ rst + m∠ vst = 180°
3 x + 7° + 9 x + 17° = 180°
12 x + 24° = 180°
12 x= 180° - 24°
12 x = 156°
x = 156° : 12
x = 13°
m ∠ rst = 3 · 13° + 7° = 39° + 7° = 46°
m ∠ vst = 180° - 46° = 134°
Answer:
A ) 46° and 134° 
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Radical functions. please help me!
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These are 8 questions and 8 answers:

1) Quesion 1:

 9+√2
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Answer: the third option:

36 + 9√7 + 4√2 + √14
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Explanation:

Multiply both numerator and denominator by the conjugate of the denominator.

The conjugate of 4 - √7 = 4 + √7

=>

\frac{9+ \sqrt{2} }{4- \sqrt{7} } . \frac{4+ \sqrt{7} }{4+ \sqrt{7} } =  \frac{(9)(4)+9 \sqrt{7}+4 \sqrt{2} + \sqrt{2} . \sqrt{7}  }{(4)^2-( \sqrt{7})^2 } =

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2) Question 2: sum

5x (\sqrt[3]{x^2y})+2( \sqrt[3]{x^5y})

Answer: fourth option

7x( \sqrt[3]{x^2y} )

Explanation:

Take x^5 out of the second radical which will result in a like term of the first radical:

5x( \sqrt[3]{x^2y} )+2( \sqrt[3]{x^5y}) =5x( \sqrt[3]{x^2y} )+2x( \sqrt[3]{x^2y})=7x( \sqrt[3]{x^2y})

which is the fourth option

3) Question 3. Which expression is equivalent to:

\frac{ \sqrt{10} }{ \sqrt[4]{8} }

Answer: the first option

Explanation

\frac{ \sqrt{10} }{ \sqrt[4]{8} } = \frac{ \sqrt[4]{10^2} }{ \sqrt[4]{8} } = \frac{ \sqrt[4]{100} }{ \sqrt[4]{8} } .  \frac{ \sqrt[4]{8^3} }{ \sqrt[4]{8^3} }  = \frac{ \sqrt[4]{(100)(512)} }{8} = \frac{ \sqrt[4]{51200} }{8} = \frac{4 \sqrt[4]{200} }{8} = \frac{ \sqrt[4]{200} }{2}

4) Question 4 What is the simplest form?

Answer: the second option

Explanation:

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5) Question 5 Product

Answer: the fourth option:

104x^4+16x^4 \sqrt{30} [/tex]\\Explanation:\\Use the square of a binomial product: (a + b)^2 = a^2 + 2ab + b^2\\[tex](4x \sqrt{5x^2} )^2+2(4x \sqrt{5x^2})(2x^2 \sqrt{6}) +(2x^2 \sqrt{6} )^2=

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6) Question 6 Product

Answer: fourth option

Explanation:

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which is the second option.

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Answer: second option y = √ (x - 2)

Explanation.

The square root function is not defined for negative values.

When x = 0, x - 2 = -2, whose square root is not defined.

Therefore, the square root of x - 2 is not defined for x = 0.
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