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dalvyx [7]
3 years ago
5

P/9 - 1 < -2 ... Solve for p

Mathematics
1 answer:
love history [14]3 years ago
7 0

Step-by-step explanation:

first collect terms, while the < remains unchange

<u>P</u><u> </u><u> </u>< 1-2

9

cross examine and get the answer

<u>P</u><u> </u>< <u>-</u><u>1</u>

9 1

<em>P</em><em> </em><em>=</em><em> </em><em>-</em><em>9</em>

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It should but 88 is still a A
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The potential energy, P, in a spring is represented using the formula P = 1/2kx2. Lupe uses an equivalent equation, which is sol
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Answer:

k=\frac{2P}{x^2}


Step-by-step explanation:

The formula is P=\frac{1}{2}kx^2

<em>We need to solve for k, so we can perform the steps shown below:</em>

P=\frac{1}{2}kx^2\\P=\frac{kx^2}{2}\\2P=kx^2\\\frac{2P}{x^2}=k


Hence, the third answer choice is correct.

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Which diagram best shows how fraction bars can be used to evaluate 1/2 divided by 1/4?
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At a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 11 m. T
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Answer:

\text{Exact area of the sidewalk}=40 \pi\text{ m}^2

\text{Approximate area of the sidewalk}=125.6\text{ m}^2

Step-by-step explanation:

We have been given that at a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 11 m. The inner edge of the sidewalk is a circle with a radius of 9 m.

To find the area of the side walk we will subtract the area of inner edge of the side walk of lion pen from the area of the outer edge of the lion pen.

\text{Area of circle}=\pi r^2, where r represents radius of the circle.

\text{Exact area of the sidewalk}=\pi*\text{(11 m)}^2-\pi*\text{(9 m)}^2

\text{Exact area of the sidewalk}=\pi*\text{121 m}^2-\pi*\text{81 m}^2

\text{Exact area of the sidewalk}=40 \pi\text{ m}^2

Therefore, the exact area of the side walk is 40 \pi\text{ m}^2

To find the approximate area of side walk let us substitute pi equals 3.14.

\text{Approximate area of the sidewalk}=40*3.14\text{ m}^2

\text{Approximate area of the sidewalk}=125.6\text{ m}^2

Therefore, the approximate area of the side walk is 125.6\text{ m}^2.

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