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Lostsunrise [7]
3 years ago
7

The equation of a line going through the point (9,2) and (-15,2)

Mathematics
1 answer:
umka21 [38]3 years ago
8 0

Answer:

y = 2

Step-by-step explanation:

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Select the factors that can be divided out to simplify the multiplication problem: 4/26 x 13/20. Select all that apply. A. 2 B.
Tcecarenko [31]
\dfrac{4}{26}  \times  \dfrac{13}{20}

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Divide by factor of 4.
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<em>*Or divide by the factor of 2 twice</em>

\dfrac{1 \times 4}{26}  \times  \dfrac{13}{5 \times 4}

\dfrac{1}{26}  \times  \dfrac{13}{5}

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Divide by factor of 13.
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\dfrac{1}{2 \times 13}  \times \ \dfrac{1 \times 13}{5}

\dfrac{1}{2}  \times  \dfrac{1 }{5}

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Combine into single fraction.
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\dfrac{1}{10}

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Answer: The factors are 2, 4 and 13
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The quotient of two negative integers results in an integer. How does the value of the quotient compared the value of the origin
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The first two were negative integers and the quotient is therefor a positive integer (Two negatives equal a positive)
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6. Which of the following is the solution set to the equation x² + 8x -15 = 5x +13?
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Answer:

The answer is A.......

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4 years ago
Write an equivalent expression using the distributive property: 7(2+4).
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(7+4) any expression 2

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The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
ankoles [38]

Answer:

     2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

  T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

  D_x=T_x\sqrt{T_x}

The mean distance of planet Y can be found using the given relation ...

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The mean distance of planet Y is increased from that of planet X by the factor ...

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2 years ago
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