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oee [108]
3 years ago
15

Please help me out I need some help

Mathematics
2 answers:
Goshia [24]3 years ago
6 0
A! since the box has 6 boxes within and only one is shaded
Vitek1552 [10]3 years ago
5 0

Answer:

A

Step-by-step explanation:

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Please help with 8 and 10 ..
Natalka [10]

Answer:

8) Parallelogram.

10) Point W.

Step-by-step explanation:

8) Give another name of plane V.

As shown the plane represents A flat shape with 4 straight sides where opposite sides are parallel and also opposite sides are equal in length.

So, plane V represents parallelogram

10) Name a point that is not coplanar with R, S and T.

Coplanar points are the points which all lie in the same plane.

So, as shown:

R, S and T are coplanar points, they all lie in the plane V

So, the point which doesn't lie in the plane V is the point W

So, <u>the point W</u> is not coplanar with R, S and T.

5 0
3 years ago
I NEED HELP ASAP<br> solve for v<br><br> v/10.24=8
11Alexandr11 [23.1K]

Answer:

v/10.24=8 V=4 <-- I am not sure if it's right but, I hope this helps u, Srry if it's the wrong answer

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What is the perimeter of the figure?
Dima020 [189]
B.
88 11/24 in 
i believe that is your answer

4 0
4 years ago
Read 2 more answers
What is the greatest common factor of <br><img src="https://tex.z-dn.net/?f=24%20%7Ba%7D%5E%7B3%7D%20c" id="TexFormula1" title="
Agata [3.3K]
If you're not sure, begin by looking for any divisor that will divide into 24a^3c and 3a without leaving a remainder.  Note that 3 is such a number, and a is another.  

Factoring out 3a from 24a^3*c and 3a, we get 3a{8a^2*c, 1}

So the GCF is 3a.

8 0
4 years ago
Find (a) the arc length and (b) the area of a sector.
Brut [27]

Answer:

a) 23.56 ft (2 dp)

b) 58.90 ft² (2 dp)

Step-by-step explanation:

<u>Formula</u>

\textsf{Arc length}=r \theta

\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta

\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}

<u>Calculation</u>

Given:

  • \theta=\dfrac{3 \pi}{2}
  • r = 5 ft

\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 5\left(\dfrac{3 \pi}{2}\right)\\& = \dfrac{15}{2} \pi \\& = 23.56\: \sf ft\:(2\:dp)\end{aligned}

\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(5^2) \left(\dfrac{3 \pi}{2}\right)\\\\& = \dfrac{25}{2}\left(\dfrac{3 \pi}{2}\right)\\\\ & = \dfrac{75}{4} \pi \\\\& = 58.90 \: \sf ft^2\:(2\:dp)\end{aligned}

6 0
3 years ago
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