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SashulF [63]
3 years ago
5

Is 35 ,000.00 in the 30,000.00 to 40,000.00 range?

Mathematics
1 answer:
Alika [10]3 years ago
3 0
It would be in the 40,000,00 range 
         Hope this helps
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Given: Two distinct circles that are tangent to each other at a common point. How many tangent lines can be drawn that touch bot
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As shown in the model below, there are three lines.

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In ΔWXY, x = 680 inches, w = 900 inches and ∠W=157°. Find all possible values of ∠X, to the nearest degree.
ikadub [295]

Given:

In ΔWXY, x = 680 inches, w = 900 inches and ∠W=157°.

To find:

The all possible values of ∠X, to the nearest degree.

Solution:

Law of Sines:

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

For ΔWXY,

\dfrac{w}{\sin W}=\dfrac{x}{\sin X}=\dfrac{y}{\sin Y}

Now,

\dfrac{w}{\sin W}=\dfrac{x}{\sin X}

\dfrac{900}{\sin (157^\circ)}=\dfrac{680}{\sin X}

900\sin X=680\sin (157^\circ)

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\sin X=0.295219

X=\sin^{-1}(0.295219)

X=17.17067

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Therefore, the value of ∠X is 17 degrees.

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3 years ago
Polygon Q is a scaled copy of Polygon P. The value of x is 6, what is the scale factor?
ANTONII [103]

Answer:

B. 3/4

Step-by-step explanation:

I just did it and got it right

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3 years ago
Shanice won 88 pieces of gum playing the bean bag toss at the county fair. That was 4 more than twice as many as Greg. How many
Andrej [43]

Answer:

42

Step-by-step explanation:

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8 0
3 years ago
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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