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marusya05 [52]
3 years ago
14

Matt is standing on top of a cliff 50 feet above a lake. The measurement of the angle of depression to a boat on the lake is 60˚

. How far is the boat from Matt?
Mathematics
1 answer:
goblinko [34]3 years ago
8 0

Answer:

28.87 feet

Step-by-step explanation:

Let the distance from Matt to the boat be x

Tan theta =opp/adj

Tan 60° = 50/ x

x = 50/Tan 60°

x = 50/1.7321

x = 28.87 feet

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It says to simplify this, by combinding like terms <br> 8 + 9x + 4 ( 11 – 2x )
erik [133]

Answer:

<h3>52+x</h3>

Step-by-step explanation:

8 + 9x + 4(11 - 2x) \\  = 8 + 9x + 44 - 8x \\  = 8 + 44 + 9x - 8x \\  = 52 + x

4 0
3 years ago
Please help! These mathematics are very confusing. need help right away.
Ostrovityanka [42]
<span>
Exercise #1:
Point H = (–2, 2)
Point J = (–2, –3)
Point K = (3, –3)

It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it.  Then
you'd immediately see what's going on.

Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.

</span><span>Notice that points J and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.

Notice that point-J is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-J is one corner of the square.

H is another corner of the square.  It's 5 units above J.

K is another corner of the square.  It's 5 units to the right of J.

The fourth corner is (2, 3) ... 5 to the right of H,
                                       and 5 above K.
____________________________________

Exercise #2:
</span><span>Point H = (6, 2)
Point J = (–2, –4)
Point K = (-2, y) .

</span><span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it.  Then
you'd immediately see what's going on.

</span><span>Notice that points J and K have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.

We need K to connect to point-H in such a way that it's on
the same horizontal line as H.  Then the vertical and horizontal
lines that meet at K will be perpendicular, and we'll have the
right angle that we need there to make the right triangle.
So K and H need to have the same y-coordinate.
H is the point (6, 2).  So K has to be up at  (2, 2) .
____________________________________________

Exercise #3:
</span>
<span>Point H = (-6, 2)
Point J = (–6, –1)
Point K = (4, 2) .
</span>
<span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it.  Then
you'd immediately see what's going on.

This exercise is exactly the same as #1, except that it's a
rectangle instead of a square.  It's still make of horizontal
and vertical lines, and that's all we need to know in order
to solve it.</span><span>

Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.

</span><span>Notice that points H and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.

Notice that point-H is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-H is one corner of the rectangle.

J is another corner of the rectangle.  It's 3 units below H.

K is another corner of the square.  It's 4 units to the right of H.

The fourth corner is (2, -1) ... 4 to the right of J,
                                       and  3 below K.

</span>
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Answer:

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