All categories

3 years ago

Each side of the Pentagon Building in Washington DC is 900 feet long. What is the perimeter ?

1 answer:

6 0

The perimeter of a Pentagon is all the length of the sides added up
A pentagon is a five sided shape
so...
900+900+900+900+900
or 900*5
That is 4500.

The perimeter of the Pentagon Building in Washington DC, whose side is 900 feet long, is 4500 feet around.

Hope this helps :)

You might be interested in

Camillo needs 2,400 oz of jelly for the food challenge. If 48 oz of jelly cost $3.84, how much will Camillo spend on jelly? Expl

pishuonlain [190]

Answer:

She'll spend 192 because 2400 divided by 48 equals 50 times that by 3.84 which will give you 192.

5 0

2 years ago

Read 2 more answers

Please help if i get this wrong i will fail

LUCKY_DIMON [66]

The answer is either b or c

3 0

3 years ago

Read 2 more answers

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

A group of 20 students and 4 adults are going on a trip to the museum. Show the ratio.

natima [27]

Answer:

5:1

Step-by-step explanation:

I think

5 0

3 years ago

Consider the polynomial function.

svp [43]

The number of positive zeros is 1. The number of negative zeros is either 3 or 1.

<h3>Rule of Descartes</h3>

This states that the number of real positive zeros of a polynomial are equal to or less than by an even number the number of sign changes of the coefficients of the polynomial, f(x). Also, the number of real negative zeros of a polynomial are equal to or less than by an even number the number of sign changes of the coefficients of the polynomial, f(-x).

<h3>The positive zero</h3>

Since f(x) = x⁴ + 2x³ - 11x² - 5x - 6

The cofficients are +1, + 2, -11, -5, -6

There is no sign change from + 1 to + 2.

There is a sign change from + 2 to - 11.

There is no sign change from - 11 to - 5.

There is no sign change from -5 to - 6.

Since there is only one sign change, there is 1 positive zero.

<h3>The negative zero</h3>

f(-x) = x⁴ - 2x³ - 11x² + 5x - 6

The cofficients are +1, - 2, -11, +5, -6

There is a sign change from + 1 to - 2.

There is no sign change from - 2 to - 11.

There is a sign change from - 11 to + 5.

There is a sign change from +5 to - 6.

Since there are 3 sign changes, we have 3 or 3 - 2 = 1 negative zeros.

So, the number of positive zeros is 1. The number of negative zeros is either 3 or 1.

Learn more about rule of descartes here:

brainly.com/question/11444977

4 0

2 years ago