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3 years ago

Find the distance between the points (3,7) and (3,2)

Mathematics

1 answer:

Serggg [28]3 years ago

3 0

Answer:

5 units.

Step-by-step explanation:

Since the distance is straight up.

You have two options:

Graph those two points and count the spaces separating them

Since they have the same x, you would only need to subtract 7-2.

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the perimeter of a rectangle swimming pool is 486 feet. The width of the pool is 35 feet less than the length. Find the length a

statuscvo [17]

Answer:

Length: 139 feet

Width: 104 feet

Step-by-step explanation:

The formula for the perimeter of a rectangle can be given by $P = 2l + 2w$ . We are given the perimeter of the pool along with the width.

$P = 486$

$w = l - 35$

From here, all we have to do is plug back into the original formula:

$486 = 2l + 2(l - 35)$

Which can be further simplified as:

$486 = 2l + 2l - 70$

$486 = 4l - 70$

From here, all we have to do is add 70 to both sides of the equation and divide by four:

$556 = 4l$

$139 = l$

To make sure that this answer is accurate, we can find that the width of the rectangle should then be 104 (given by 139 - 35). All we have to do is plug back into the original equation:

$P = 2l + 2w$

$P = 2(139) + 2(104)$

$P = 278 + 208$

$P = 486$

And the substitution works, so the length of the rectangle would be 139 feet and the width would be 104 feet.

5 0

3 years ago

Read 2 more answers

P(n) = n + 1; Find p(6)

True [87]

Answer:

p(6) = 7

Step-by-step explanation:

p(n) = n + 1

Let n = 6

p(6) = 6 + 1

p(6) = 7

7 0

3 years ago

Read 2 more answers

Simplify.<br><br> x8 ÷ x4 · x2

miskamm [114]

<span>x^8 ÷ x^4 · x^2
= </span>x^4 · x^2
= x^6

hope it helps

5 0

3 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago

A central angle in a circle with radius 7 inches has the measure θ , in radians. The length of the arc subtended by the central

Ahat [919]

The relationship between arc, radius, and central angle in radians is:

Θ = s/r

So just plugging in the values:

Θ = (5π/3)/7 = 5π/21 ≈ 0.748 or 0.7

6 0

3 years ago