(-9,-13) , [7 1/2, 3 1/2], (5,1), (4,0), (0,__) what is the next number?

AveGali [126]

Answer:

x = 155

Step-by-step explanation:

First find the missing angle in the smaller triangle.

We know that the sum of the angles in a triangle is 180 so 35 + 90 + the missing angle = 180 so 180 - (35+90) = 180 - 125 = 55

Since the line above is a straight line, the sum of the angles must equal 180 so 55 + 60 + ? = 180. So 180 - 115 = 65.

The triangle is a right triangle so that means on of the angles is 90. So 65 + 90 = 155, so we have 180 - 155 = 25.

The angle that x is on is also a straight line so that means the angles must add up to 180 so 180 - 25 = 155.

8 0

3 years ago

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Find the area in the shaded region.

irinina [24]

The answer should be 227 ft

7 0

3 years ago

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Relationship B has a lesser rate than Relationship A. The graph represents Relationship A.

Zinaida [17]

Given that Relationship B has a lesser rate than Relationship A and that the graph representing Relationship A is a first-quadrant graph showing a ray from the origin through the points (2, 3) and (4, 6) where the horizontal axis label is Time in weeks and the vertical axis label is Plant growth in inches.

The rate of relationship A is given by the slope of the graph as follows:

$slope= \frac{6-3}{4-2} = \frac{3}{2} =1.5$

To obtain which table could represent Relationship B, we check the slopes of the tables and see which has a lesser slope.

For table A.
Time (weeks) 3 6 8 10
Plant growth (in.) 2.25 4.5 6 7.5

$slope= \frac{4.5-2.25}{6-3} = \frac{2.25}{3} =0.75$

For table B.
Time (weeks) 3 6 8 10
Plant growth (in.) 4.8 9.6 12.8 16

 $slope= \frac{9.6-4.8}{6-3} = \frac{4.8}{3} =1.6$

 For tabe C.
Time (weeks) 3 4 6 9
Plant growth (in.) 5.4 7.2 10.8 16.2

 $slope= \frac{7.2-5.4}{4-3} = \frac{1.8}{1} =1.8$

For table D.
Time (weeks) 3 4 6 9
Plant growth (in.) 6.3 8.4 12.6 18.9

 $slope= \frac{8.4-6.3}{4-3} = \frac{2.1}{1} =2.1$ 

Therefore, the table that could represent Relationship B is table A.

7 0

4 years ago

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I need help with finding the answer to a) and b). Thank you!

shtirl [24]

Answer:

$\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }$

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}$

$\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(x)=-\frac{21}{29}$

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

$a=\sqrt{29^2-21^2}=20$

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

$180$

So:

$\displaystyle 90 < \frac{x}{2} < 135$

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

1)

Recall that:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}$

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}$

Simplify:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

2)

Likewise:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }$

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }$

Simplify:

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

3)

Finally:

$\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}$

Therefore:

$\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}$

5 0

3 years ago

Which option below correctly describes the domain of (see image) and explains why the range of b is not all real numbers?

Reika [66]

The domain is all real numbers less than or equal to 81.

The range is not all real numbers because the square root symbol applied to a non-negative number is non-negative

7 0

3 years ago

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