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Klio2033 [76]
3 years ago
10

The perimeter of a semicircle is 25.7 feet. What is the semicircle's radius? Use 3.14 for ​.

Mathematics
1 answer:
jarptica [38.1K]3 years ago
7 0

Answer: 4.09

Step-by-step explanation:

25.7 divided by pie is 8.18 which is the diameter. divide that by 2 and u get the radius.

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Monica measures 91 milliliters of water into 9 tiny beakers. she measures an equal amount of water into the first 8 beakers. she
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To determine the amount of water in the first 8 beakers, you will subtract the 19 ml that are in the 9th beaker from the total to see how much water was actually put into the first 8 beakers.

91 ml - 19 ml = 72 ml.

The 72 ml are divided evenly between 8 beakers.

72/8 = 9 ml

There would be 9 ml of water in each of the first 8 beakers.
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Answer:

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A water balloon has a diameter of 6 inches. How many cubic inches of water does it contain when it is 3/4 full
Travka [436]
Below are the choices that can be found from other sources:

A) 36 pi 
<span>B) 6 pi </span>
<span>C) 27 pi </span>
<span>D) 216 pi </span>
<span>E) 48 pi
</span>
A sphere's volume is (4/3)*pi*r^3. 
<span>The diameter of the balloon is 6 inches. The diameter is twice the radius. The radius is 3 inches. </span>

<span>(4/3)*pi*3^3 = 36 * pi </span>

<span>But the balloon is only 3/4 full. So </span>

<span>36 * (3/4) * pi = 27 * pi. </span>

<span>The balloon has 27 pi cubic inches of water.</span>
7 0
3 years ago
use the information provided to write the standard form equation of each hyperbola vertices:(4,14), (4,-10) foci: (4,15), (4,-11
Evgen [1.6K]
So... if you notice the picture below, based on the given vertices, is a hyperbola with a vertical traverse axis

meaning for the equation, the fraction with the "y" variable is the positive fraction, thus \bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1

so... what' the point h,k for the center?  well, just take a peek at the graph, the center is half-way between both vertices, that's h,k

what's the "a" component or the traverse axis... well, is the distance from one vertex to the center, notice the picture, notice how many units from either vertex to the center, that's "a"

now.. what's "b"?  well, "b" comes from the conjugate axis, or the other runnning over the x-axis   hmm the smaller one in this case.... well, we dunno what "b" is

however, we know the distance from either focus, to the center of the hyperbola, and that distance "c", is  \bf c=\sqrt{a^2+b^2}

notice the picture, notice the distance from either focus to the center, that's the distance "c", thus  \bf c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b

that'd be "b"

bearing in mind that   \bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1&#10;\qquad &#10;\begin{cases}&#10;center= ({{ h}},{{ k}})\\\\ b=\sqrt{c^2-a^2}&#10;\end{cases}

8 0
3 years ago
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