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Alinara [238K]
3 years ago
12

There are 98,515 tress at Washington Park.If there are 85 acres of land,and the trees spread out evenly,how many trees are there

on each acre of land?
(show work )
Mathematics
2 answers:
torisob [31]3 years ago
6 0

Answer

Find out the how many trees are there on each acre of land .

To prove

Let us assume that the trees are there on each acre of land = x

As given

There are 98,515 tress at Washington Park.

If there are 85 acres of land

the trees spread out evenly i.e the tree are equally divided in each acre of land .

Than the equation becomes

x = \frac{98515}{85}

x = 1159

Therefore are 1159 trees on each acre of land .

Hence proved

aksik [14]3 years ago
3 0
To answer this all we need to do is DIVIDE.

98515 ÷ 85 = <span>1159

CHECK OUR WORK

1159 </span>× 85 = 98515
<span>
So, we were right!! :) 

So, there would be 1159 trees on each acre.

Hope I helped ya!!

</span>
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Inga [223]

Answer:

Simplify the numerator with the sine expression by taking A = B = x and simplify the denominator using A = x and B = 2x. The resulting denominator will have sin (2x) and cos (2x) which you can write in terms of sin (x) and cos (x) using the expressions above.

Step-by-step explanation:

4 0
2 years ago
: Which equations have an infinite number of solutions? Select all that apply. A -2(x - 2) + 3x = x - 4 B 2x + 4(x - 1) = 3(2x +
MissTica

Option C and Option D

3x + 2(x - 2) + 6 = 2(2x + 3) + x - 4 has infinite number of solutions

4(x + 2) + x + 1 = 2x - 3 + 3(x + 4) has infinite number of solutions

<h3><u>Solution:</u></h3>

Given that we have to find the equations that has infinite number of solutions

If we end up with the same term on both sides of the equal sign, such as 4 = 4 or 4x = 4x, then we have infinite solutions.

If we end up with different numbers on either side of the equal sign, as in 4 = 5, then we have no solutions.

<h3><u>Option A</u></h3>

-2(x - 2) + 3x = x - 4

Multiply the terms inside the bracket

-2x + 4 + 3x = x - 4

x + 4 = x - 4

Thus this equation does not have infinite number of solutions

<h3><u>Option B</u></h3>

2x + 4(x - 1) = 3(2x + 1) - 2(x - 1)

2x + 4x - 4 = 6x + 3 - 2x + 2

6x - 4 = 4x + 5

6x - 4x = 5 + 4

2x = 9

x = \frac{9}{2}

Thus this equation has only one solution.

Thus this equation does not have infinite number of solutions

<h3><u>Option C</u></h3>

3x + 2(x - 2) + 6 = 2(2x + 3) + x - 4

3x + 2x - 4 + 6 = 4x + 6 + x - 4

5x + 2 = 5x + 2

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<h3><u>Option D</u></h3>

4(x + 2) + x + 1 = 2x - 3 + 3(x + 4)

4x + 8 + x + 1 = 2x - 3 + 3x + 12

5x + 9 = 5x + 9

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3 years ago
I need help with questions #7 and #8 plz
katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

c^2 = a^2 + b^2 - 2ab \cos C

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

7.

We use the law of cosines to find C.

18^2 = 12^2 + 16^2 - 2(12)(16) \cos C

324 = 144 + 256 - 384 \cos C

-384 \cos C = -76

\cos C = 0.2

C = \cos^{-1} 0.2

C = 78.6^\circ

Now we use the law of sines to find angle A.

Law of Sines

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

We know c and C. We can solve for a.

\dfrac{a}{\sin A} = \dfrac{c}{\sin C}

\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}

Cross multiply.

18 \sin A = 12 \sin 78.6^\circ

\sin A = \dfrac{12 \sin 78.6^\circ}{18}

\sin A = 0.6535

A = \sin^{-1} 0.6535

A = 40.8^\circ

To find B, we use

m<A + m<B + m<C = 180

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m<B = 60.6 deg

8.

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

a^2 = b^2 + c^2 - 2bc \cos A

b^2 = a^2 + c^2 - 2ac \cos B

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A = 20.7^\circ

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b^2 = a^2 + c^2 - 2ac \cos B

18^2 = 8^2 + 12^2 - 2(8)(12) \cos B

324 = 208 - 192 \cos A

\cos B = -0.6042

B = 127.2^\circ

Find angle C:

c^2 = a^2 + b^2 - 2ab \cos C

12^2 = 8^2 + 18^2 - 2(8)(18) \cos B

144 = 388 - 288 \cos A

\cos C = 0.8472

C = 32.1^\circ

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