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3 years ago

Answer quickly please! Find the perimeter of the figure to the nearest hundredth.

Mathematics

1 answer:

Ronch [10]3 years ago

5 0

10+8+6=24

π x 12/2
6 π
24+6 π= 42.84955592

Answer= 42.849cm

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Number of ponits = number of games won x3 + number of games drawn Grangers have got 42 points. They have drawn 6 games. How many

Helga [31]

Answer:

Number of games Grangers won = 12

Step-by-step explanation:

Given:

Number of ponits = [Number of games won]3 + number of games drawn

Grangers have = 42 points

Number of game drawn = 6 games

Find:

Number of games Grangers won

Computation:

Number of ponits = [Number of games won]3 + number of games drawn

42 = [Number of games won][3] + 6

36 = [Number of games won][3}

[Number of games won] = 12

Number of games Grangers won = 12

7 0

3 years ago

What is the calculator display for 5/12

Dmitry_Shevchenko [17]

5/12 = 0.4166666...

That is what it says on the calculator.

8 0

4 years ago

Read 2 more answers

-3r²+11t-6 factoring trinomials

IgorC [24]

- (3r squared - 11t + 6)

4 0

3 years ago

Suppose astronomers built a 50-meter telescope. how much greater would its light-collecting area be than that of the 10-meter ke

drek231 [11]

Answer: Its light-collecting area would be 25 times greater than that of the 10-meter keck telescope.

Step-by-step explanation:

1. To solve this problem you must apply the formula for calculate the area of a circle, which is shown below:

$A=r^{2}\pi$

Where<em> r</em> is the radius of the circle.

2. The diameter of a circle is:

$D=\frac{r}{2}$

Where <em>r</em> is the radius of the circle.

3. Therefore, keeping this on mind, you have that the light-collecting area of a 50-meter keck telescope is:

$A_1=(\frac{50m}{2})^{2}\pi=1963,49m^{2}$

4. And the light-collecting area of a 10-meter keck telescope is:

$A_2=(\frac{10m}{2})^{2}\pi=78.53m^{2}$

5. Divide $A_1$ by $A_2$ , then:

$\frac{1963.49m^{2}}{78.53m^{2}}=25$

6. Therefore, its light-collecting area would be 25 times greater than that of the 10-meter keck telescope.

4 0

4 years ago

For a circle of radius 3 feet, find the arc lengths subtended by a central angle of 57 degrees.

tigry1 [53]

$\bf \textit{arc's length}\\\\
s=\cfrac{\theta \pi r}{180}\quad 
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad degrees\\
------\\
r=3\\
\theta = 57
\end{cases}\implies s=\cfrac{57\cdot \pi \cdot 3}{180}$

7 0

3 years ago

Read 2 more answers