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Ber [7]
3 years ago
7

need help with this it is due at 11:59 answer correctly ad will mark you brainless and you dont need to show the work just the a

nswer
Mathematics
1 answer:
katen-ka-za [31]3 years ago
3 0
The picture is blank sorry:/
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A circle has a radius of 9 inches. The Radius is multiplied by 2/3 to form a second circle. How is the ratio of the areas relate
liraira [26]

Answer:

\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{r_{1} }{r_{2}}) ^{2}

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii.

Step-by-step explanation:

Radius of first circle (r_{1}) = 9 inches

Area of first circle = \pi r_{1} ^{2}

Area of first circle = 9 × 9 × π = 81 π

Now, since the radius is multiplied by 2/3 for from a new circle.

∴ Radius of the second circle = 9 \times \frac{2}{3} = 6\ inches

Area of second circle =  \pi r_{2} ^{2}

Area of second circle = 6 × 6 × π = 36 π

Now,

\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81\pi }{36\pi }

\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{9}{6}) ^{2} = (\frac{r_{1} }{r_{2}}) ^{2}

∵ (r_{1}) = 9 inches and (r_{2}) = 6 inches

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii. i.e., \frac {radius\ of\ first\ circle)^{2} }{(radius\ of\ second\ circle)^{2} } = \frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)}

8 0
3 years ago
What is 3/5 of 3 3/4
murzikaleks [220]
\dfrac{3}{5}\cdot3\dfrac{3}{4}=\\
\dfrac{3}{5}\cdot\dfrac{15}{4}=\\
3\cdot\dfrac{3}{4}=\\
\dfrac{9}{4}
5 0
3 years ago
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Answer the question please I don’t get it either
Sophie [7]

Answer:

The top angle is 60 degrees

Step-by-step explanation:

Triangle angle sum equals to 180. 90 plus 30 is 120 and when you subtract that from 180 tada! 60

3 0
3 years ago
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What is the equation of the graph
Katen [24]
I assume you want it in slop-intercept form.
y = 3/4x + y-intercept (the graph isn't labeled)
If you just want to find the slope of the graph, it would be 3/4
6 0
3 years ago
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H(x)= (x^2 -36)/x +6 explain why the following function is not continuous at x=-6
sergeinik [125]

You can solve this problem through factoring.

First, you have the equation,

h(x) = \frac{x^2-36}{x-6}

Then, you can factor the numerator.

h(x) = \frac{(x+6)(x-6)}{x-6}

You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.

You are left with h(x) = x+6

The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.

8 0
3 years ago
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