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maw [93]
3 years ago
6

Find the volume of the sphere to the nearest whole number. Use pi = 3.14. A. 131 in. 3 B. 393 in. 3 C. 4,187 in. 3 D. 523 in. 3

Mathematics
1 answer:
Westkost [7]3 years ago
5 0
The answer is B. That's what I got
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You are graphing y < 2x + 1. What type of line do you use and where do you shade?
bazaltina [42]
Since there isn't a line under the < sign, this means that we used a dotted or dashed line. The dotted or dashed line indicates that we do NOT include the boundary as part of the solution set. 

Since y is isolated and we have a less than sign, this means we shade below the dashed/dotted boundary line. Specifically, the boundary line is the graph of y = 2x+1. This boundary line goes through (0,1) and (1,3). Again, points on this boundary line are NOT part of the solution set. 

So in summary we have:
A dashed or dotted boundary line
The shaded region is below the dashed/dotted boundary line.
3 0
4 years ago
5x + 2x + 2 = 9x + 4​
bija089 [108]

Step-by-step explanation:

Given

5x + 2x + 2 = 9x + 4

9x - 5x - 2x = 2 - 4

9x - 7x = - 2

2x = - 2

x = - 2 / 2

Therefore x = -1.

5 0
3 years ago
Read 2 more answers
If a/\2+1/a/\2=11, then a+1/a=?
Nadya [2.5K]

Answer:

\pm \sqrt{13}

Step-by-step explanation:

\bigg(a +  \frac{1}{a}  \bigg)^{2} =  {a}^{2}   +  \frac{1}{ {a}^{2} }  + 2 \times a \times  \frac{1}{a}  \\  \\ \bigg(a +  \frac{1}{a}  \bigg)^{2} = 11 + 2 \\  \\   \bigg(a +  \frac{1}{a}  \bigg)^{2} = 13 \\  \\  \bigg(a +  \frac{1}{a}  \bigg) =  \pm \sqrt{13}

5 0
3 years ago
2. The long-jump pit was recently rebuilt to make it level with the runway. Volunteers provided pieces of
lakkis [162]

Answer:2. It’s 24.58

How many boards did the volunteers supply round your calculations to the nearest hundred- 13.44

How many whole boards-14

Step-by-step explanation:

8 0
3 years ago
After a rotation of 90° about the origin, the coordinates of the vertices of the image of a triangle are A'(6, 3), B'(-2, 1),
Rasek [7]

Answer:

<em>A</em>(-3, 6), <em>B</em>(-1, -2), <em>C</em>(-7, 1)

Step-by-step explanation:

To the pre-image after a 270°-counterclockwise rotation [90°-clockwise rotation], just reverse it by doing a 270°-clockwise rotation [90°-counterclockwise rotation]:

Extended Rotation Rules

  • 270°-clockwise rotation [90°-counterclockwise rotation] >> (x, y) → (-y, x)
  • 270°-counterclockwise rotation [90°-clockwise rotation] >> (x, y) → (y, -x)
  • 180°-rotation >> (x, y) → (-x, -y)

So, perform your rotation:

270°-clockwise rotation [90°-counterclockwise rotation] → <em>C</em><em>'</em>[1, 7] was originally at <em>C</em>[-7, 1]

→ <em>B'</em>[-2, 1] was originally at <em>B</em>[-1, -2]

→ <em>A</em><em>'</em>[6, 3] was originally at <em>A</em>[-3, 6]

I am joyous to assist you anytime.

3 0
3 years ago
Read 2 more answers
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