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Ivan
3 years ago
13

Lindsay draws a right triangle and adds the measures of the right angle and one acute angle. Which is a possible sum of the two

angles?
Mathematics
2 answers:
Anon25 [30]3 years ago
6 0
Anything between 91 degrees and 179 degrees. You can pick any between these two.
Viktor [21]3 years ago
4 0

Answer:

90< \gamma< 180

Step-by-step explanation:

An acute angle is an angle that measures more than 0º and less than 90º. And a right angle is an angle that measures exactly 90º. Thus:

Let:

\alpha = Right\hspace{3}angle=90^{\circ}\\\beta= Acute\hspace{3}angle\hspace{10}0^{\circ}

So, in this sense, the sum of the two angles can´t be equal or greater than 180º, since \beta , also it has to be at least greater than 90º since \beta>0^{\circ}

Therefore the sum of the two angles is:

90^{\circ}< \gamma< 180^{\circ}

In another words, the sum is greater than 90º and less than 180º

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Answer:

My guess is 21

Step-by-step explanation:

4 0
3 years ago
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A right cylinder has a radius of r inches and height of 2r inches.
lys-0071 [83]

Answer:

Part a) The lateral area is 4r^{2} \pi \ in^{2}

Part b) The area of the two bases together is 2r^{2} \pi\ in^{2}

Part c) The surface area is 6r^{2} \pi\ in^{2}

Step-by-step explanation:

we know that

The surface area of a right cylinder is equal to

SA=LA+2B

where

LA is the lateral area

B is the area of the base of cylinder

we have

r=r\ in

h=2r\ in

Part a) Find the lateral area

The lateral area is equal to

LA=2\pi rh

substitute the values

LA=2\pi r(2r)

LA=4r^{2} \pi\ in^{2}

Part b) Find the area of the two bases together

The area of the  base B is equal to

B=r^{2} \pi\ in^{2}

so

the area of the two bases together is

2B=2r^{2} \pi\ in^{2}

Part c) Find the surface area of the cylinder

SA=LA+2B

we have

LA=4r^{2} \pi\ in^{2}

2B=2r^{2} \pi\ in^{2}

substitute

SA=4r^{2} \pi+2r^{2} \pi=6r^{2} \pi\ in^{2}

4 0
3 years ago
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Need help for this question TYSM MWA®
xz_007 [3.2K]

Answer:

1/6y+1/6(y+12)-2... so the third one

Step-by-step explanation:

1/6y+1/6(y+12)-2

1/6y+1/6y+2-2

2/6y

Reduces to 1/3y

4 0
3 years ago
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Find the value of y and x.
max2010maxim [7]

Answer:

  • x = 130°
  • y = 65°

Step-by-step explanation:

All inscribed angles intercepting the same arc have the same measure.

If you slide the vertex of angle y clockwise around the circle until it coincides with the vertex of the angle marked 65°, you see that those two inscribed angles (y and 65°) intercept the same arc, so have the same measure:

  y = 65°.

The measure of an inscribed angle is half the measure of the intercepted arc, so the arc marked x is double the angle 65°.

  x = 130°.

6 0
3 years ago
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The box plots show the high temperatures in June and August for Denver in degrees Fahrenheit.
Kryger [21]

Hi There!

---------------------------------------------------------

Full Question:

The box plots show the high temperatures in June and August for Denver in degrees Fahrenheit.

Which can you tell about the mean temperatures for these two months?

There is not enough information to determine the mean temperatures.

The mean temperature for August is higher than June's mean temperature.

The mean temperature for June is equal to the mean temperature for August.

The high interquartile range for August pulls the mean temperature above June's mean temperature.

---------------------------------------------------------

Interquartile Range Formula: Q3 - Q1

Interquartile Range for August: 10

Interquartile Range for June: 8

---------------------------------------------------------

Median = Mean

June: 82

August: 82

---------------------------------------------------------

Answer: The mean temperature for June is equal to the mean temperature for August.

---------------------------------------------------------

Hope This Helps :)

5 0
3 years ago
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