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Ierofanga [76]
1 year ago
11

An athlete ran 800 meters in 160 seconds. What was his rate in meters per minute? Enter your answer in the box.

Mathematics
1 answer:
FromTheMoon [43]1 year ago
3 0

The athlete's average speed was 300 meters/minute.

<h3>How to calculate the speed of the athlete?</h3>

To calculate the speed of the athlete we must perform the following operations.

Divide the distance into the total time:

  • 800m ÷ 160sec = 5m/sec

Transform the value from seconds to minutes, for which we must multiply 5 by 60 because each minute is made up of 60 seconds.

  • 5m/sec × 60sec = 300m/min

According to the above, the average speed of the athlete is 300m/min.

Learn more about speed in: brainly.com/question/7359669

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For the given equation;

(3x-5)^2=-125

We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.

Let us begin by expanding the parenthesis;

\begin{gathered} (3x-5)^2\Rightarrow(3x-5)(3x-5) \\ (3x-5)(3x-5)=9x^2-15x-15x+25 \\ (3x-5)^2=9x^2-30x+25 \end{gathered}

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\begin{gathered} 9x^2-30x+25=-125 \\ \text{Add 125 to both sides and we'll have;} \\ 9x^2-30x+25+125=-125+125 \\ 9x^2-30x+150=0 \end{gathered}

We shall now solve the resulting quadratic equation using the quadratic formula as follows;

\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}

ANSWER:

x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3}

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Also, the corresponding angles of similar triangles are congruent.

Therefore, The angles of ΔA'B'C are congruent to the corresponding parts of the original triangle.

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