He following diagram, the ship is anchored to the ocean floor. The length of the rope attaching the

Mathematics

1 answer:

mr_godi [17]2 years ago

4 0

The distance between the boat and the ocean's floor is 19m

<h3>How to get the distance from the ship to the ocean's floor?</h3>

We can see this as a right triangle, where the rope with the anchor is the hypotenuse. So we already know that the hypotenuse measures 30m.

We also know an angle, it measures 39°, and we want to get the opposite cathetus to said angle (the height).

Then we use the relation:

Sin(a) = (opposite cathetus)/(hypotenuse).

Replacing the values that we know, we get:

Sin(39°) = d/30m

sin(39°)*30m = d = 18.9m ≈ 19m

The distance between the boat and the ocean's floor is 19m.

If you want to learn more about right triangles, you can read:

brainly.com/question/2217700

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A company manufactures a 14-ounce box of cereal. Boxes are randomly weighed to ensure the correct amount. If the discrepancy in

Stolb23 [73]

Answer:

Answer is B on Edge

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Step-by-step explanation:

8 0

3 years ago

Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)

Mnenie [13.5K]

One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

so
$(x+2)^2+(y-3)^2=18$
$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
$(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2$

the radius is $3\sqrt{2}$
center is (-2,3)

find distance between (8,4) and (-2,3)

$D=\sqrt{(8-(-2))^2+(4-3)^2}$
$D=\sqrt{(8+2)^2+(1)^2}$
$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

$r=3\sqrt{2}$ ≈4.2
$D=\sqrt{101}$ ≈10.04

do r<D

(8,4) is outside the circle

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3 years ago

A rectangle has length that is 5 less than 2 times the width. Write an expression for the area of the rectangle.

a_sh-v [17]

We are given a relationship between the sides of a rectangle, that is, the length of one of its sides is 5 less two times its width, and we are asked to find an expression for the area. Let's remember that the area of a rectangle is equal to the product of the length of its side by its width. Let "w" be the length of the rectangle and "L" its lenght, then the area is given by the following formula: