# The Pentagon in Washington DC is immense in size, with each outside wall 921 feet in length. What is the perimeter of the Pentagon?

QUESTION POSTED AT 05/12/2019 - 10:05 AM

QUESTION POSTED AT 05/12/2019 - 10:05 AM

The perimeter of the Pentagon would be 4,605 feet. Add 921, 5 times because a pentagon has 5 sides. When you add 921, 5 times, you should get 4,605 feet.

This can be answered through simple ratio and proportion

Let x = width

y = height

x/y = x2/y2 = x3/y3

since

for 18 inches width we have 26 inches as length, therefore

18/26 = 45/y2

y2 = 65 ft

there is no need to convert the units for the smaller dimensions since we are dealing with definite ratios thus unitless.

ANSWERED AT 18/01/2020 - 01:12 PM

QUESTION POSTED AT 18/01/2020 - 01:12 PM

We let

r1 = the original radius of the cylinder

r2 = the new radius of the cylinder

h1 = the original height of the cylinder

h2 = the new height of the cylinder

SA = surface are of the cylinder

the radius and the height are shrunk down to a third of their value therefore

r2 = r1/3

h2 = h1/3

SA = 2(PI)r^2 + 2(PI)rh

SA = 2(PI)[(r1^2)/9] +2(PI)(r1/3)(h1/3)

Simplifying

SA = [2(PI)r]/9 * (r1 + h1)

ANSWERED AT 18/01/2020 - 01:12 PM

QUESTION POSTED AT 18/01/2020 - 01:12 PM

ANSWERED AT 18/01/2020 - 01:07 PM

QUESTION POSTED AT 18/01/2020 - 01:07 PM

**Answer: Yes, the side lengths form a pythagorean triple.**

**Step-by-step explanation: **Given that the side lengths of triangle ABC are as follows :

We are to **check whether the side lengths form a pythagorean triple or not.**

**For making a pythagorean triple, the side lengths must satisfy the following :**

**We have**

**So,**

**Thus, the side lengths will form a pythagorean triple.**

ANSWERED AT 18/01/2020 - 01:06 PM

QUESTION POSTED AT 18/01/2020 - 01:06 PM

Area of a square is length squared.

Let the length be x, then

Therefore,

Let the length be x, then

Therefore,

ANSWERED AT 18/01/2020 - 01:05 PM

QUESTION POSTED AT 18/01/2020 - 01:05 PM

ANSWERED AT 18/01/2020 - 12:57 PM

QUESTION POSTED AT 18/01/2020 - 12:57 PM

32.5? that could be an accuret answer

ANSWERED AT 18/01/2020 - 12:55 PM

QUESTION POSTED AT 18/01/2020 - 12:55 PM

16x2−4x−4x+1=4x(4x−1)−1(4x−1)=(4x−1)(4x−1)=(4x−1)2
Length of each side=4x-1

ANSWERED AT 18/01/2020 - 12:54 PM

QUESTION POSTED AT 18/01/2020 - 12:54 PM

Its a scalene triangle if it is a triangle

ANSWERED AT 18/01/2020 - 12:53 PM

QUESTION POSTED AT 18/01/2020 - 12:53 PM

D-C-E

DC = 6m - 21

CE = 7m - 18

DE = 65

DC + CE = DE

6m - 21 + 7m - 18 = 65

13m - 39 = 65

13m = 65 + 39

13m = 104

m = 104/13

m = 8

DC = 6m - 21 = 6(8) - 21 = 48 - 21 = 27

CE = 7m - 18 = 7(8) - 18 = 56 - 18 = 38

DC = 6m - 21

CE = 7m - 18

DE = 65

DC + CE = DE

6m - 21 + 7m - 18 = 65

13m - 39 = 65

13m = 65 + 39

13m = 104

m = 104/13

m = 8

DC = 6m - 21 = 6(8) - 21 = 48 - 21 = 27

CE = 7m - 18 = 7(8) - 18 = 56 - 18 = 38

ANSWERED AT 18/01/2020 - 12:52 PM

QUESTION POSTED AT 18/01/2020 - 12:52 PM

Get the perimeter of the each shape and add it all up to get the total fencing material required.

2 square flower beds = 2 x 4a = 8a

1 rectangular play ground = 2l + 2w

total fencing material = 8a + 2l + 2w

2 square flower beds = 2 x 4a = 8a

1 rectangular play ground = 2l + 2w

total fencing material = 8a + 2l + 2w

ANSWERED AT 18/01/2020 - 12:50 PM

QUESTION POSTED AT 18/01/2020 - 12:50 PM

192=1/2 x b x 12

192=6b

b=32mm

192=6b

b=32mm

ANSWERED AT 18/01/2020 - 12:48 PM

QUESTION POSTED AT 18/01/2020 - 12:48 PM

**Answer:**

No triangles can be constructed

**Step-by-step explanation:**

we know that

The Triangle Inequality Theorem, states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

so

In this problem

------> is not true

therefore

No triangles can be constructed with the given side lengths

ANSWERED AT 18/01/2020 - 12:46 PM

QUESTION POSTED AT 18/01/2020 - 12:46 PM

Answer – 19 inches

The smallest possible whole-number length of the longest side of an obtuse triangle is the next integer greater than the Pythagorean distance of the other two sides.

Given,

A = 10

B = 15

Pythagoras theorem;

C^2 = A^2 + B^2

C^2 = 15^2 + 10^2

C^2 = 225 + 100

C^2 = 325

C = 18.03

Therefore, the smallest possible whole-number length of the longest side is the next integer greater than 18.03 inches.

The smallest possible whole-number length of the longest side is 19 inches

The smallest possible whole-number length of the longest side of an obtuse triangle is the next integer greater than the Pythagorean distance of the other two sides.

Given,

A = 10

B = 15

Pythagoras theorem;

C^2 = A^2 + B^2

C^2 = 15^2 + 10^2

C^2 = 225 + 100

C^2 = 325

C = 18.03

Therefore, the smallest possible whole-number length of the longest side is the next integer greater than 18.03 inches.

The smallest possible whole-number length of the longest side is 19 inches

ANSWERED AT 18/01/2020 - 12:45 PM

QUESTION POSTED AT 18/01/2020 - 12:45 PM

The first part is already in squared terms. "19.25 sq ft"
Taking the measurements for the mirror 3ft and 5ft, you'd get: "15 sq ft."
Subtract that answer from the previous number to get "4.25 sq ft.

Hope this helps out a little lol :)

Hope this helps out a little lol :)

ANSWERED AT 18/01/2020 - 12:38 PM

QUESTION POSTED AT 18/01/2020 - 12:38 PM

The length is 112 because 16*7=112

ANSWERED AT 18/01/2020 - 12:38 PM

QUESTION POSTED AT 18/01/2020 - 12:38 PM

The answer is 4 3/4 yards of the fabric.

This can be calculated using the proportion.

If the 3 1/6 yards of the fabric is enough for 1 1/3 shirts, how many yards are necessary of 2 shirts:

Let's express 3 1/6 as 19/6:

Similarly, 1 1/3 = 4/3:

This can be calculated using the proportion.

If the 3 1/6 yards of the fabric is enough for 1 1/3 shirts, how many yards are necessary of 2 shirts:

Let's express 3 1/6 as 19/6:

Similarly, 1 1/3 = 4/3:

ANSWERED AT 18/01/2020 - 12:37 PM

QUESTION POSTED AT 18/01/2020 - 12:37 PM

| x - (-30) |

which can be simplifed into

| x + 30 |

We use absolute value because distance cannot be negative.

Laurie must stay within 10 feet. Therefore, the distance x is from -30 feet has to be equal to 10 to find the maximum and minimum depths that Laurie can stay between:

| x + 30 | = 10

When we solve this absolute value equation, we work with two cases.

Case 1: When x + 30 is positive, then |x+30| = x+30 (no change) and we have the equation

x + 30 = 10

Case 2: When x + 30 is negative, then |x+30| = -(x+30) so as to force the distance into a positive number. Then this gives us the equation

-(x+30) = 10

which we can multiply both sides by -1 to get

x + 30 = -10

ANSWERED AT 18/01/2020 - 12:35 PM

QUESTION POSTED AT 18/01/2020 - 12:35 PM

It could be alot the easest is 20° and 20°

ANSWERED AT 18/01/2020 - 12:32 PM

QUESTION POSTED AT 18/01/2020 - 12:32 PM

The answer is 141.35 ft²

Before the first break, it was painted:

150 ft² ÷ 2 = 75 ft²

Now it's left:

150 ft² - 75 ft² = 75 ft²

Before the second break, it was painted:

75 ft² ÷ 2 = 37.5 ft²

Now it's left:

75 ft² - 37.5 ft² = 37.5 ft²

Before the third break, it was painted:

37.5 ft² ÷ 2 = 18.75 ft²

Now it's left:

37.5 ft² - 18.75 ft² = 18.75 ft²

Before the fourth break, it was painted:

18.75 ft² ÷ 2 = 9.375 ft²

Now it's left:

18.75 ft² - 9.375 ft² = 9.375 ft²

Before the fourth break, it was painted:

9.375 ft² ÷ 2 = 4.6875 ft²

Now it's left:

9.375 ft² - 4.6875 ft² = 4.6875 ft²

Now, we will sum what he painted for now:

75 ft² + 37.5 ft² + 18.75 ft² + 9.375 ft² 4.6875 ft² = 141.3125 ft² ≈ 141.35 ft²

When the painter takes his fifth break, there will be 141.35 ft² of the wall painted.

Before the first break, it was painted:

150 ft² ÷ 2 = 75 ft²

Now it's left:

150 ft² - 75 ft² = 75 ft²

Before the second break, it was painted:

75 ft² ÷ 2 = 37.5 ft²

Now it's left:

75 ft² - 37.5 ft² = 37.5 ft²

Before the third break, it was painted:

37.5 ft² ÷ 2 = 18.75 ft²

Now it's left:

37.5 ft² - 18.75 ft² = 18.75 ft²

Before the fourth break, it was painted:

18.75 ft² ÷ 2 = 9.375 ft²

Now it's left:

18.75 ft² - 9.375 ft² = 9.375 ft²

Before the fourth break, it was painted:

9.375 ft² ÷ 2 = 4.6875 ft²

Now it's left:

9.375 ft² - 4.6875 ft² = 4.6875 ft²

Now, we will sum what he painted for now:

75 ft² + 37.5 ft² + 18.75 ft² + 9.375 ft² 4.6875 ft² = 141.3125 ft² ≈ 141.35 ft²

When the painter takes his fifth break, there will be 141.35 ft² of the wall painted.

ANSWERED AT 18/01/2020 - 12:28 PM

QUESTION POSTED AT 18/01/2020 - 12:28 PM

Well if you have a congruent lines, and you have P = 50 and xz = 15
P = xz + 2 * wz so to find wz you need to subtract xz and divide by 2
50 - 15 = 35
and then divide by 2 to get 17.5

ANSWERED AT 18/01/2020 - 12:26 PM

QUESTION POSTED AT 18/01/2020 - 12:26 PM

If it was more details, it would be easier to solve. But I can suggest a logical way. ZB + BA = ZA. If B is the midpoint it devides line on two equal lines.

ANSWERED AT 18/01/2020 - 12:25 PM

QUESTION POSTED AT 18/01/2020 - 12:25 PM

Hello,

4x²-12x+9=(2x)²-2*(2x)*3+3²=(2x-3)²

4x²-12x+9=(2x)²-2*(2x)*3+3²=(2x-3)²

ANSWERED AT 18/01/2020 - 12:24 PM

QUESTION POSTED AT 18/01/2020 - 12:24 PM

6 feet by 10 feet would use 60 feet of fencing. that's the biggest I can think of without going over 62 feet

ANSWERED AT 18/01/2020 - 12:15 PM

QUESTION POSTED AT 18/01/2020 - 12:15 PM

Given:

62 feet of fencing = perimeter

P = 2 (L + W)

62/2 = 31

16 + 15 = 16 * 15 = 240 ft²

17 + 14 = 17 * 14 = 238 ft²

18 + 13 = 18 * 13 = 234 ft²

19 + 12 = 19 * 12 = 228 ft²

20 + 11 = 20 * 11 = 220 ft²

The best dimensions that will give the greatest area is to have a length of 16 feet and a width of 15 feet.

62 feet of fencing = perimeter

P = 2 (L + W)

62/2 = 31

16 + 15 = 16 * 15 = 240 ft²

17 + 14 = 17 * 14 = 238 ft²

18 + 13 = 18 * 13 = 234 ft²

19 + 12 = 19 * 12 = 228 ft²

20 + 11 = 20 * 11 = 220 ft²

The best dimensions that will give the greatest area is to have a length of 16 feet and a width of 15 feet.

ANSWERED AT 18/01/2020 - 12:15 PM

QUESTION POSTED AT 18/01/2020 - 12:15 PM

**Answer:**

The answer is square pyramid.

**Step-by-step explanation:**

All of the following solid figures except a **square pyramid** have two bases.

The square pyramid has four side faces that are triangles. The base of square pyramid is a square and it has five vertices with eight edges.

The cube has two congruent flat bases and four flat faces. This means, we can say, there are a total of 6 flat faces in a cube.

The pentagonal prism has also two bases. And the cylinder also has 2 bases.

ANSWERED AT 18/01/2020 - 12:14 PM

QUESTION POSTED AT 18/01/2020 - 12:14 PM

Let us assume the width of the rectangle = w

Let us assume the length of the rectangle = l

Then

l = 2w

Also

Perimeter of the rectangle = 2 (l + w)

24 = 2 (2w + w)

24 = 2 (3w)

24 = 6w

w = 24/6

= 4 inches

Now

The length of the rectangle = 2w

= 2 * 4 inches

= 8 inches

So the length of the rectangle is 8 inches and the width is 4 inches.

Let us assume the length of the rectangle = l

Then

l = 2w

Also

Perimeter of the rectangle = 2 (l + w)

24 = 2 (2w + w)

24 = 2 (3w)

24 = 6w

w = 24/6

= 4 inches

Now

The length of the rectangle = 2w

= 2 * 4 inches

= 8 inches

So the length of the rectangle is 8 inches and the width is 4 inches.

ANSWERED AT 18/01/2020 - 12:13 PM

QUESTION POSTED AT 18/01/2020 - 12:13 PM

Given:

square shaped traffic sign. side measure of 8x - 2

A square has 4 equal sides. To find the area of a square, simply square the measure of 1 side.

Area of a square = a²

A = (8x-2)²

A = (8x-2)(8x-2)

A = 8x(8x-2) - 2(8x-2)

A = 64x² - 16x - 16x + 4

**A = 64x² - 32x + 4**

square shaped traffic sign. side measure of 8x - 2

A square has 4 equal sides. To find the area of a square, simply square the measure of 1 side.

Area of a square = a²

A = (8x-2)²

A = (8x-2)(8x-2)

A = 8x(8x-2) - 2(8x-2)

A = 64x² - 16x - 16x + 4

ANSWERED AT 18/01/2020 - 12:13 PM

QUESTION POSTED AT 18/01/2020 - 12:13 PM

**Answer:**

x = 3; the diver hits the water 3 feet away horizontally from the board

**Step-by-step explanation:**

ANSWERED AT 18/01/2020 - 12:08 PM

QUESTION POSTED AT 18/01/2020 - 12:08 PM

To get the perimeter of the quadrilateral, the distance formula can be used. We simply have to subtract the y coordinate of one point to another and the same for the x coordinate. The square root of the sum of the squares of the difference is the distance.The results can then be added to get the perimeter. So,

CF = sqrt ( (2-(-1))^2 + (-1-(-2))^2 ) = sqrt(10)

FE = sqrt ( (-2-1)^2 + (2-1)^2 ) = sqrt(10)

ED = sqrt ( (-1-(-2))^2 + (-2-2)^2 ) = sqrt(17)

DC = sqrt ( (2-(-1)^2 + (-1-(-2))^2 ) = sqrt(10)

P = CF + FE + ED + DC = sqrt(10) + sqrt(10) + sqrt(17) + sqrt(10) = 13.61

CF = sqrt ( (2-(-1))^2 + (-1-(-2))^2 ) = sqrt(10)

FE = sqrt ( (-2-1)^2 + (2-1)^2 ) = sqrt(10)

ED = sqrt ( (-1-(-2))^2 + (-2-2)^2 ) = sqrt(17)

DC = sqrt ( (2-(-1)^2 + (-1-(-2))^2 ) = sqrt(10)

P = CF + FE + ED + DC = sqrt(10) + sqrt(10) + sqrt(17) + sqrt(10) = 13.61

ANSWERED AT 18/01/2020 - 11:59 AM

QUESTION POSTED AT 18/01/2020 - 11:59 AM