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3 years ago

On a treasure map, a X is placed at point (9, -7). if the pirates are currently on point (9,3), how many units away are they?

Mathematics

2 answers:

Alex_Xolod [135]3 years ago

8 0

Answer:

They are 10 units away from the treasure

Step-by-step explanation:

Notice that the points (9, -7) and (9, 3) are both located at the same x-value (9) while their y-values are in one case 3 units above the x-axis, and in the other 7 units BELOW the x-axis. Therefore thy differ by 3 - (-7) units , which totals 10 units.

kozerog [31]3 years ago

4 0

The answer to this question is 10 units away

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Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified l

Sloan [31]

Answer:

The integral of the volume is:

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

The result is: $V = 78.97731$

Step-by-step explanation:

Given

Curve: $x^2 + 4y^2 = 4$

About line $x = 2$ --- Missing information

Required

Set up an integral for the volume

$x^2 + 4y^2 = 4$

Make x^2 the subject

$x^2 = 4 - 4y^2$

Square both sides

$x = \sqrt{(4 - 4y^2)$

Factor out 4

$x = \sqrt{4(1 - y^2)$

Split

$x = \sqrt{4} * \sqrt{(1 - y^2)$

$x = \±2 * \sqrt{(1 - y^2)$

$x = \±2 \sqrt{(1 - y^2)$

Split

$x_1 = -2 \sqrt{(1 - y^2)}\ and\ x_2 = 2 \sqrt{(1 - y^2)}$

Rotate about x = 2 implies that:

$r = 2 - x$

So:

$r_1 = 2 - (-2 \sqrt{(1 - y^2)})$

$r_1 = 2 +2 \sqrt{(1 - y^2)}$

$r_2 = 2 - 2 \sqrt{(1 - y^2)}$

Using washer method along the y-axis i.e. integral from 0 to 1.

We have:

$V = 2\pi\int\limits^1_0 {(r_1^2 - r_2^2)} \, dy$

Substitute values for r1 and r2

$V = 2\pi\int\limits^1_0 {(( 2 +2 \sqrt{(1 - y^2)})^2 - ( 2 -2 \sqrt{(1 - y^2)})^2)} \, dy$

Evaluate the squares

$V = 2\pi\int\limits^1_0 {(4 +8 \sqrt{(1 - y^2)} + 4(1 - y^2)) - (4 -8 \sqrt{(1 - y^2)} + 4(1 - y^2))} \, dy$

Remove brackets and collect like terms

$V = 2\pi\int\limits^1_0 {4 - 4 + 8\sqrt{(1 - y^2)} +8 \sqrt{(1 - y^2)}+ 4(1 - y^2) - 4(1 - y^2)} \, dy$

$V = 2\pi\int\limits^1_0 { 16\sqrt{(1 - y^2)} \, dy$

Rewrite as:

$V = 16* 2\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

Using the calculator:

$\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy = \frac{\pi}{4}$

So:

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

$V = 32\pi * \frac{\pi}{4}$

$V =\frac{32\pi^2}{4}$

$V =8\pi^2$

Take:

$\pi = 3.142$

$V = 8* 3.142^2$

$V = 78.97731$ --- approximated

3 0

3 years ago

A box of pancake mix states that 1 1/2 cups of pancake mix and 1 cup of water are needed to make 8 pancakes. How many cups of pa

lilavasa [31]

Answer:

I think your answer is B

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Corey has 16 cups of flour to make cookies. One batch of cookies takes 2 1/2 cups of flour. If he must save 6 cups of flour for

White raven [17]

16-6=10
2.5x=10
Divide by 2.5
X=4
So Corey can make 4 batches of cookies

8 0

3 years ago

a rectangular garden is 30 ft by 40 ft. Part of the garden is removed in order to install a walkway of uniform width around it.

Stels [109]

Answer:

The wide of the walkway is 5ft.

Step-by-step explanation:

In first place you have a rectangular garden of 30ft by 40 ft as shown in Figure 1 (Added file).

So, the area of this rectangle is:

$Area of rectangle = width * length = 30ft * 40ft =1200ft$

Area=1200ft

Now, the new garden has a walkway of uniform width around it, and the area of the new rectangle is one-half the area of the old garden

$Area of the new garden= \frac{1200}{2} =600ft$

In order to be a walkway of uniform width it's necessary to substract of each side the same amount, for this case will be 5 ft, as shown in the Figure 2. Now the new rectangule is 20ft by 30ft.

$Areaofthenewrectangle=20ft*30ft=600ft$