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Tju [1.3M]
3 years ago
14

Can you help me with #7

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
7 0
2 miles-12 min.
3 miles- 18min
4 miles-24 min
5 miles-30 min

3 miles- 18
4 miles-24
5 miles-30
6 miles-36

Yes they are the same
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Y ∝ <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D" id="TexFormula1" title="\frac{1}{x}" alt="\frac{1}{x}" align="absmi
SashulF [63]
It should be your answer i think
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3 years ago
Convert the degree measurement to radians. Express answer as multiple of π: 126°
almond37 [142]

Answer:

.7 pi    Radians

Step-by-step explanation:

There are pi radians in a 180 degrees :

   pi / 180   * 126  = .7 pi  Radians

8 0
2 years ago
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Can someone add these up and tell me what my final grades would be
denis23 [38]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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A circular pond 24 yd in
Schach [20]

Answer:

The walk will cost $8164.

Step-by-step explanation:

Given:

Diameter of the circular pond (D) = 24 yd

Width of the gravel path (x) = 2 yd

Cost per yard of the path = $50

Now, radius of the circular pond is half of the diameter and is given as:

Radius,R=\frac{D}{2}=\frac{24}{2}=12\ yd

Now, area of the pond is given as:

A_{pond}=\pi R^2=3.14\times (12)^2=3.14\times 144=452.16\ yd^2

Area of the complete path including the pond area is given as:

A_{outer}=\pi(R+x)^2=3.14\times(12+2)^2=3.14\times196=615.44\ yd^2

Now, area of the gravel path can be obtained by subtracting the pond area from the total outer area. This gives,

A_{path}=A_{outer}-A_{pond}\\\\A_{path}=615.44-452.16=163.28\ yd^2

Now, using unitary method,

Cost of 1 square yard of path = $50

∴ Cost of 163.28 square yard of path = 50 × 163.28 = $8164

Hence, the walk will cost $8164.

5 0
3 years ago
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Please answer these questions! will mark brainlist!
timofeeve [1]

<u><em>1</em><em>s</em><em>t</em><em> </em><em>photo....</em></u>

<em>Yes</em>

<em>Explanation</em>

All when divided becomes 1/3 , so it is proportional...

<em>Pls </em><em>mark</em><em> </em><em>brainliest</em>

<u><em>2</em><em>n</em><em>d</em><em> </em><em>photo</em><em>.</em><em>.</em><em>.</em><em>.</em></u>

Yes<em> </em>

<em>Explai</em><em>n</em><em>ation</em>

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