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Gala2k [10]
3 years ago
15

Solve with work please

Mathematics
1 answer:
Irina18 [472]3 years ago
3 0

Answer:

150 yd

Step-by-step explanation:

Since the line segment is an angle bisector then it divides the opposite side into segments that are proportional to the other 2 sides , that is

\frac{250}{375} = \frac{x}{225} ( x is the distance from the hole to the oak tree )

Cross- multiply

375x = 56250 ( divide both sides by 375 )

x = 150

The hole is 150 yards from the oak tree.

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Answer:

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Step-by-step explanation:

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Can anyone help me on this?
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Which point lies in the solution set of the inequality 1/2(x+2)+3y&lt;8?
nasty-shy [4]

Note: Options are missing. So, the general solution of the given inequality is shown below.

Given:

The inequality is:

\dfrac{1}{2}(x+2)+3y

To find:

The point lies in the solution set of the given inequality.

Solution:

We have,

\dfrac{1}{2}(x+2)+3y

It can be written as:

\dfrac{1}{2}(x)+\dfrac{1}{2}(2)+3y

\dfrac{1}{2}x+1+3y

\dfrac{1}{2}x+3y

\dfrac{1}{2}x+3y

All the points that satisfy the above inequality are in the solution set of the given inequality.

For example (0,0).

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0

This statement is true. It means (0,0) is in the solution set.

For example (0,3).

\dfrac{1}{2}(0)+3(3)

9

This statement is false. It means (0,3) is not in the solution set.

The graph of the inequality \dfrac{1}{2}x+3y is shown below.

All the points in the shaded region are in the solution set but the points on the boundary line are not in the solution set.

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For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim
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Answer:

Part 1) A=36(\pi-2)\ cm^2

Part 2) P=6(\pi+2\sqrt{2})\ cm

Step-by-step explanation:

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)

we have that the base and the height of triangle is equal to the radius of the circle

r=12\ cm\\b=12\ cm\\h=12\ cm

substitute

A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2

simplify

Factor 36

A=36(\pi-2)\ cm^2

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r

substitute the given values

C=\frac{1}{2}\pi (12)\\C=6\pi\ cm

The hypotenuse of right triangle is equal to (applying the Pythagorean Theorem)

AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm

simplify

AC=12\sqrt{2}\ cm

Find the perimeter

P=(6\pi+12\sqrt{2})\ cm

simplify

Factor 6

P=6(\pi+2\sqrt{2})\ cm

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