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Nataly [62]
2 years ago
8

(06.04 HC) A figure is located at (2,0),(2,-2), and (6,0) on a coordinate plan. What kind of 3-D shape would be created if the f

igure was rotate around the x-axis? Provide an explanation and proof of your answer to receive full credit.Include the dimensions of the 3-D shape in your explanation.(10 points)
Mathematics
1 answer:
fiasKO [112]2 years ago
6 0

Answer:

a cone

Step-by-step explanation:

When you plot the three points on the graph it makes a triangle. As it rotates around the x axis, the points at 2,0 and 2,-2 make a circle, which becomes the base of the cone.

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PLEASE HELP ASAP WILL GIVE BRAINLEIST!!!
lyudmila [28]

Answer:

$ 3125.00

Step-by-step explanation:

P=\frac{SI * 100 }{R*T}=

\frac{500*100 }{4*4} = \frac{50000}{16} = $3125.00

7 0
3 years ago
What is the value of x? <br><br> Enter your answer in the box.<br><br> X =
melamori03 [73]

Answer:

x=54

Step-by-step explanation:

2x + 2 = 3x - 52

subtract 2x from both sides, now u have:

2 = x - 52

add 52 from both sides, now u have

54 = x

7 0
3 years ago
Write a division problem whose quotient has its first digit in the hundreds place
Katen [24]
<span>Here's one --> 1000÷100=10</span>
6 0
3 years ago
Please help?
Mazyrski [523]

Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0
3 years ago
4
emmasim [6.3K]

Answer:

A.    

Step-by-step explanation:

y=mx + b where m is the slope and b is the y-intercept.  So substituting 4/5 for m and 5 for b, you get answer A.

4 0
3 years ago
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