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Andrew [12]
1 year ago
7

Find the approximate side length of a square game board with an area of 105 in2.

Mathematics
1 answer:
aleksley [76]1 year ago
8 0

Answer:

Rounded to the nearest inch: 10

Rounded to the nearest tenth: 10.2

Rounded to the nearest hundredth: 10.25

Step-by-step explanation:

To find the area of a square, we multiply the side length by itself, or in other words, SQUARE the side length!

A=s^2

We are given the area, so plugging that into our equation looks like this:

105=s^2

To eliminate the exponent, take the square root of both sides of the equation!

\sqrt{105}=s

From here, there are a few paths to take, depending on what your teacher is looking for. I will evaluate the approximate of the square root by finding the nearest perfect squares.

The nearest perfect square less than √105 is √100, and the nearest perfect square greater than √105 is √121.

105 is closer to 100 than to 121, so √105 is closer to √100.

√100 is equal to 10, so the √105 is approximately 10 inches!

Using a calculator to check:

\sqrt{105}=10.247 inches

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Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then \frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }.  By the Mean Value Theorem, there is a number c such that 0 < c with v'(c)=60 \:{\frac{mi}{h^2}}. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 60 \:{\frac{mi}{h^2}}.

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

f'(c)=\frac{f(b)-f(a)}{b-a}

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then v(0 \:h) = 30 \:{\frac{mi}{h} } and v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} } (note that 20 minutes is 20/60=1/3 of an hour), so the average rate of change of v on the interval [0 \:h, \frac{1}{3} \:h] is

\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in (0 \:h, \frac{1}{3} \:h) at which v'(c)=60 \:{\frac{mi}{h^2}}.

c is a time time between 2:00 and 2:20 at which the acceleration is 60 \:{\frac{mi}{h^2}}.

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snow_lady [41]
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Answer:

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