MATHEMATICS MIDDLE SCHOOL

Write each of the numbers 1, 4, 9, 16, and 25 as a base raised to the second power. Explain why these numbers are sometimes called "perfect squares"

Answers

Answer 1

Answer:

1 as a base raised to the second power is $1^2 = 1$
4 as a base raised to the second power is $2^2 = 4$
9 as a base raised to the second power is $3^2 = 9$
16 as a base raised to the second power is $4^2 = 16$
25 as a base raised to the second power is $5^2 = 25$
They are perfect squares because their square roots are whole numbers.

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A number x as a base raised to the second power is written as:

$x = √(x)^2$

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1, 4, 9, 16, and 25

We find the square root of each of these numbers, so:

Since $√(1) = 1$ , 1 as a base raised to the second power is $1^2 = 1$ .

Since $√(4) = 2$ , 4 as a base raised to the second power is $2^2 = 4$

Since $√(9) = 3$ , 9 as a base raised to the second power is $3^2 = 9$

Since $√(16) = 4$ , 16 as a base raised to the second power is $4^2 = 16$

Since $√(26) = 2$ , 25 as a base raised to the second power is $5^2 = 25$

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Each number that has a whole square root is called perfect square, which is the case for 1, 4, 9, 16, and 25.

A similar question is found at brainly.com/question/2675310

Answer 2

Answer: First, we can raise each number to the second power by multipying it by itself. :)
1^2=1
4^2=16
9^2=81
16^2=256

These numbers are called "perfect squares" because their square roots are whole numbers, rather than decimals. For example, If you had a square with an area of 16, the side legnths of the square would be the whole (thus "perfect") number 4. For this reason, 16 (4^2) is considered a "perfect square" number. I hope that makes sense!! :)

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The paths of two runners cross at a stop sign. One runner is heading south at a constant rate of 6.5 miles per hour while the other runner is heading west at a constant rate of 7 miles per hour. How fast is the distance between them changing after 10 minutes?

Answers

Answer:

The distance between them changing after 10 minutes will be 9.553 mph.

Step-by-step explanation:

The paths of two runners cross at a stop sign (O). One runner is heading south at a constant rate of 6.5 miles per hour towards A while the other runner is heading west at a constant rate of 7 miles per hour towards B.

So, after 10 minutes the first runner covers a distance of $6.5 * (10)/(60) = 1.083$ miles and the second runner covers a distance of $7 * (10)/(60) = 1.167$ miles.

Therefore, after 10 minutes their distance will be $\sqrt{(1.083)^(2) + (1.167)^(2)} = 1.592$ miles.

Now, the distance between them is given by

AB² = OA² + OB²

Now, differentiating this equation with respect to time t (in hours) we get

$2(AB) (d(AB))/(dt) = 2(OA) (d(OA))/(dt) + 2(OB) (d(OB))/(dt)$

⇒ $(AB) (d(AB))/(dt) = (OA) (d(OA))/(dt) + (OB) (d(OB))/(dt)$

⇒ $1.592 * (d(AB))/(dt) = 1.083 * 6.5 + 1.167 * 7 = 15.208$

⇒ $(d(AB))/(dt) = 9.553$ mph.

Therefore, the distance between them changing after 10 minutes will be 9.553 mph. (Answer)

Final answer:

The distance between the two runners is not changing after 10 minutes.

Explanation:

To find the rate of change of the distance between the two runners, we can use the concept of relative velocity. The distance is changing due to the motion of both runners, so we need to find the rate at which each runner is approaching or moving away from the other. Since one runner is heading south and the other is heading west, their velocities are perpendicular to each other. We can use the Pythagorean theorem to find their combined velocity and then calculate the rate of change of the distance between them.

Let's consider the southward runner as Runner A and the westward runner as Runner B. The velocity of A is 6.5 miles per hour, and the velocity of B is 7 miles per hour. After 10 minutes, the distance traveled by A can be calculated as (6.5 miles/hour) * (10/60) hours = 1.083 miles. The distance traveled by B can be calculated as (7 miles/hour) * (10/60) hours = 1.167 miles.

Using the Pythagorean theorem, we can calculate the distance between the two runners after 10 minutes:

Distance = sqrt((1.083 miles)^2 + (1.167 miles)^2) ≈ 1.563 miles

To find the rate of change of the distance between them, we can differentiate the equation for the distance with respect to time:

d(Distance)/dt = (1/2)*((2*(1.083 miles)*(0))/(sqrt((1.083 miles)^2 + (1.167 miles)^2))) + (1/2)*((2*(1.167 miles)*(0))/(sqrt((1.083 miles)^2 + (1.167 miles)^2))) = 0

Therefore, the distance between the two runners is not changing after 10 minutes.