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3 years ago

How do you find the area of a circle if you already know the diameter

Mathematics

2 answers:

kramer3 years ago

7 0

Answer:

Area = Pi x Radias^2

aev [14]3 years ago

4 0

Answer:

A=πr2

Step-by-step explanation:

radius or (r) is the diameter in half. so you can take this formula and find the area

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An icicle it's in the shape of an inverted cone with a diameter of 12 m m and a height of 100 mm. How much Frozen water is in th

krek1111 [17]

Answer:

3768 mm ^ 3

Step-by-step explanation:

We have that the volume of a cone is given by:

V = 1/3 * Ac * h

Where Ac is the area of the circle, we know that the radius is half the diameter then:

r = d / 2

r = 12/2

r = 6

And Ac is equal to:

Ac = pi * r ^ 2

replacing:

Ac = 3.14 * 6 ^ 2

Ac = 113.04

113.04 mm ^ 2 is the area of the circle, replacing the volume form knowing that h = 100

V = 1/3 * 113.04 * 100

V = 3768

Therefore they fit a total of 3768 mm ^ 3

4 0

3 years ago

Please help! It's due today.

Georgia [21]

Answer:

Step-by-step explanation:

First substitute the numbers for x and y: -6(-9)-3(-6)+2(-9)

(the screenshot below will explain the rest)

7 0

3 years ago

Read 2 more answers

A male student of the author has a measured pulse rate of 52 beats per

Liono4ka [1.6K]

Answer:

$z=-1.49$

Step-by-step explanation:

$\text{Standard Score, z} =\dfrac{X-\mu}{\sigma} $ where:\\\\Mean Pulse rate, \mu =67.3$ beats per minute\\Standard Deviation, \sigma = 10.3$ beats per minute.\\$

For a male student who has a measured pulse rate of 52 beats per minute.

Raw Score, X =52 beats per minute.

Therefore:

$\text{Standard Score, z} =\dfrac{52-67.3}{10.3}\\z=-1.49$

Since the usual pulse rates are within 2 standard deviations of the mean, a z-score of -1.49 tells us that the selected student's pulse rate is within the usual pulse rates.

8 0

3 years ago

11.5,9.4, ,5.2 what is the blank

Alla [95]

The answer is 7.3...Subtract by 2.1

3 0

2 years ago

Read 2 more answers

Find the height of a prism with a base in the form of a sector in which the central angle is 25 degree . its volume is found to

jasenka [17]

V=BH
$B=A_{sector}= \frac{r^2\pi\alpha}{360^o} 
\\H=4r
\\350= \frac{r^2\pi\times 25^o}{360^o}4r
\\350= \frac{r^3\pi\times 100^o}{360^o}
\\ r^3\times 3.14\times 100^o=350\times 360^o
\\r^3= \frac{350\times 360^o}{3.14\times 100^o} 
\\r= \sqrt[3]{\frac{350\times 360^o}{3.14\times 100^o} } 
\\H=4r=4 \sqrt[3]{\frac{350\times 360^o}{3.14\times 100^o} } \approx 29.5$

6 0

2 years ago