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3 years ago

10

according to the Greek mathematician Zeno, if each bounce of a bakl is half the height of the bounce begire it, the ball will ne

ver stop bouncing. write the fractions in hundredths that should be written at points B and C

2 answers:

aev [14]3 years ago

8 0

I can't answer because I don't know where B or C is!!!

USPshnik [31]3 years ago

6 0

You have to say what or where point B and C is for me to answer it

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A triangle has one side that is 5 units long and an adjacent angle

Alexxx [7]

Answer:

<u>you can get help with this link given</u>

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2 years ago

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Help this is extremely confusing

Sati [7]

Answer:

x = 10

Step-by-step explanation:

With UW extended to point X, then

∠VWX is an external angle of the triangle.

The external angle of a triangle is equal to the sum of the 2 opposite interior angles, that is

∠VWX = ∠WUV + ∠UVX ← substitute values

6x - 19 = 3x + 3 + x - 2, that is

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3 years ago

[100 POINTS+BRAINLIEST] Use the figure to find each missing angle measure.

devlian [24]

Answer:

Angle 1 is 112°, 2 is 68°, 3 is 90°, 4 is also 90°, 5 is 22° and angle 6 is 158°

Step-by-step explanation:

To find angles 3 &4 and 2 &1, you subtract the measurement given in each intersection from 180 (all straight lines are 180) to find the other angle measurements. To find 5 I used the 4th angle and the 2nd angle to find the missing number out of 180 since all of the angles in a triangle have a sum of 180°. The missing angle was 22. You can use the angle measurement of #5 to find 6 like how I mentioned before about all straight lines equaling 180°. If this all sounds like mumbo-jumbo I can elaborate a little more in the comment section!

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matrenka [14]

tax because the tax shows you what you paid and ect

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3 years ago

The area of a mirror is 225 Square inches, it's width is

Zepler [3.9K]

NO. The mirror will not fit in a space that is 15 inches by 16 inches

<em><u>Solution:</u></em>

Given that area of mirror is 225 square inches

$width = 13\frac{3}{4} \text{ inches }$

Converting the above mixed fraction we get,

$width = 13\frac{3}{4} = \frac{13 \times 4 + 3}{4} = \frac{55}{4} \text{ inches }$

Let us find the length of mirror

The area of mirror is given as:

area of mirror = length x width

Substituting the given values,

$225 = length \times \frac{55}{4}\\\\length = 225 \times \frac{4}{55}\\\\length = 225 \times 0.0727\\\\length = 16.36$

Thus length of mirror is 16.36 inches

<em><u>Will the mirror fit in a space that is 15 inches by 16 inches?</u></em>

NO. The mirror will not fit in a space that is 15 inches by 16 inches

Because length of mirror is 16.36 inches whereas the given space is 15 inches long

16.36 > 15 so the length of mirror will not fit inside the space

Also width of mirror is $\frac{55}{4} = 13.75$ inches which is less than the given space whose width is 16 inches

13.75 < 16 so the width of mirror will not fit inside the space

7 0

3 years ago