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3 years ago

15

How many sides does a regular polygon have with each exterior angle measures up to 30

1 answer:

ch4aika [34]3 years ago

5 0

The answer would be 6 exteriors.

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Last tuesday a doughnuts store sold 20 bavarian cream, 60 glazed, 15 chocolate

Murljashka [212]

Answer:

Step-by-step explanation:

b:g:c:r = 20:60:15:25 = 4:12:3:5

b:c = 4:3

7 0

2 years ago

If de has endpoints D(-1,6)and E(3,-2),then the midpoint M of DE lies in Quadrant1

SVETLANKA909090 [29]

$M(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) \\M(\frac{-1 + 3}{2}, \frac{6 + (-2)}{2}) \\ M(\frac{2}{2}, \frac{6 - 2}{2}) \\ M(1, \frac{4}{2}) \\ M(1, 2)$

7 0

3 years ago

777dan777 [17]

Answer:

$\begin{array}{ccc}\text{Radius}&\text{Volume of sphere}&\text{Volume of cylinder}\\&&\\1&\dfrac{4}{3}\pi &2\pi \\&&\\2&\dfrac{32}{3}\pi &16\pi \\&&\\3&36\pi &54\pi \\&&\\4&\dfrac{256}{3}\pi &128\pi \\&&\\5&\dfrac{500}{3}\pi &250\pi\end{array}$

Step-by-step explanation:

Use formulas for the volumes:

$V_{sphere}=\dfrac{4}{3}\pi r^3,\\ \\V_{cylinder}=\pi r^2h=\pi r^2\cdot 2r=2\pi r^3.$

1. When r=1,

$V_{sphere}=\dfrac{4}{3}\pi\cdot 1^3=\dfrac{4}{3}\pi,\\ \\V_{cylinder}=2\pi \cdot 1^3=2\pi.$

2. When r=2,

$V_{sphere}=\dfrac{4}{3}\pi\cdot 2^3=\dfrac{32}{3}\pi,\\ \\V_{cylinder}=2\pi \cdot 2^3=16\pi.$

3. When r=3,

$V_{sphere}=\dfrac{4}{3}\pi\cdot 3^3=36\pi,\\ \\V_{cylinder}=2\pi \cdot 3^3=54\pi.$

4. When r=4,

$V_{sphere}=\dfrac{4}{3}\pi\cdot 4^3=\dfrac{256}{3}\pi,\\ \\V_{cylinder}=2\pi \cdot 4^3=128\pi.$

5. When r=5,

$V_{sphere}=\dfrac{4}{3}\pi\cdot 5^3=\dfrac{500}{3}\pi,\\ \\V_{cylinder}=2\pi \cdot 5^3=250\pi.$

Note that for all r,

$\dfrac{V_{sphere}}{V_{cylinder}}=\dfrac{\frac{4}{3}\pi r^3}{2\pi r^3}=\dfrac{2}{3}.$

8 0

3 years ago

Read 2 more answers

FAST PLEASE!!!!! 10 POINTS!!!!!!! SHOW WORK FOR ALL!!!!!!

Dimas [21]

Step-by-step explanation:

-6 ( x - 2y ) ( -12x ) / ( -2 )

-6 ( -3 - 2*-5 ) -12 / -2 = 6

-6 ( -3 + 10) 6x

-6 ( 7 )

-42

6 0

3 years ago

A particle moves along the x-axis so that at time t≥0t\ge 0t≥0 its position is given by x(t)=−t3+6t2−9t+42.x(t)=-t^3+6t^2-9t+42.

riadik2000 [5.3K]

Answer: 4 units

Step-by-step explanation:

The given position function: $x(t)=-t^3+6t^2-9t+42.$ , where t≥0.

To determine: Total distance traveled by the particle from 0≤ t ≤ 4.0.

At t=0,

$x(0)=-(0)^3+6(0)^2-9(0)+42=42\\$

At, t=4,

$x(4)=-(4)^3+6(4)^2-9(4)+42\\\\=-64+6(16)-36+42=38$

Total distance traveled by the particle from 0≤ t ≤ 4.0 = |42-38| units

= 4 units

Hence, the distance traveled by the particle from 0≤t≤4.0 is 4 units.

4 0

2 years ago