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Ludmilka [50]
3 years ago
13

Please help me this is very important due today

Mathematics
1 answer:
alexdok [17]3 years ago
4 0

Answer:

I'd say D) a worried young man

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Write a fraction that is less than 5/6 and has a denominator of 8
KatRina [158]
You'll want to make a common denominator with 6 and 8.
That denominator would be 24.
24/6=4 so you would have to multiply 5/6 by 4/4 to get 20/24.
Next, 24/8=3 so 1/8 could be multiplied by 3/3 to get 3/24.
Since 3 is less than 20, 1/8 is smaller than 5/6.

If you want the same numerator, 5/8 = 15/24. This would make 5/8 smaller than 5/6 as well.
The highest eighth you can go is 6/8 which is 18/24.
So you can use any numerator between 1 and 6 with a denominator of 8 to get a fraction smaller than 5/6.
3 0
3 years ago
The figure shows a circle graphed on a coordinate plane
frosja888 [35]
I'm pretty sure that you need to multiply pie times the radius of the circle
4 0
4 years ago
Write 342 to 1 significant figure​
Rufina [12.5K]

Answer:

300

Step-by-step explanation:

A significant figure is the most important (largest) number you can round it to.

As it wants 1 significant figure, you count 1 to the left and round the 4 down.

Hope this helps :)

5 0
3 years ago
How big is the earth
melomori [17]
The 4th largest planet
7 0
3 years ago
Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level. how m
GalinKa [24]

The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

For Addison h=85(1/3)

d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)

Subtracting both distances we get

8\sqrt{2}-6\sqrt{2}  =2\sqrt{2}

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

brainly.com/question/1392973

5 0
3 years ago
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