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3 years ago

How do you calculate the area of a circle?

Mathematics

2 answers:

never [62]3 years ago

6 0

Answer:

$Area of circle = \pi (radius)^2 =\pi (r)^2$

Step-by-step explanation:

We need to find about how to calculate the area of a circle.

And match with the given choices. Where given choices are:

Area of a circle is equal to the diameterdiameter

squaredradiusradius squared times diameterheightpiwidth.

I'm not sure what exactly you have typed lol :)

But I can still answer that.

You just need to apply formula of the area of circle which is given by:

$Area of circle = \pi (radius)^2 =\pi (r)^2$

mote1985 [20]3 years ago

6 0

Answer:

radius squared times pi

Step-by-step explanation:

edge

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Week 1 2 3 4 5 Water Level (inches) − 1 1 4 1 3 8 2 1 2 − 1 5 8 − 1 3 4 Between which two weeks did the water level change the m

suter [353]

Answer:

The water level change the most in week 3 and week 4

The change = -370 inches

Step-by-step explanation:

Given - Week 1 2 3 4 5

Water Level (inches) − 1 1 4 1 3 8 2 1 2 -1 5 8 -1 3 4

To find - Between which two weeks did the water level change the most? Calculate the change .

Proof -

The formula for change in water level with respect to week be-

Rate of Change in Water Level = Difference in water level / Difference in weeks

Now,

For week 1 and week 2

Rate of change in water level = $\frac{138 - 114}{2 - 1}$ = $\frac{24}{1}$ = 24 inches

Now,

For week 1 and week 3

Rate of change in water level = $\frac{212 - 114}{3 - 1}$ = $\frac{98}{2}$ = 49 inches

Now,

For week 1 and week 4

Rate of change in water level = $\frac{-158 - 114}{4 - 1}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 1 and week 5

Rate of change in water level = $\frac{-134 - 114}{5 - 1}$ = $-\frac{248}{4}$ = -62 inches

Now,

For week 2 and week 3

Rate of change in water level = $\frac{212 - 138}{3 - 2}$ = $\frac{74}{1}$ = 74 inches

Now,

For week 2 and week 4

Rate of change in water level = $\frac{-158 - 138}{4 - 2}$ = $-\frac{296}{2}$ = -148 inches

Now,

For week 2 and week 5

Rate of change in water level = $\frac{-134 - 138}{5 - 2}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 3 and week 4

Rate of change in water level = $\frac{-158 - 212}{4 - 3}$ = $-\frac{370}{1}$ = -370 inches

Now,

For week 3 and week 5

Rate of change in water level = $\frac{-134 - 212}{5 - 3}$ = $-\frac{346}{2}$ = 173 inches

Now,

For week 4 and week 5

Rate of change in water level = $\frac{-134 + 158}{5 - 4}$ = $\frac{24}{1}$ = 24 inches

∴ we get

The water level change the most in week 3 and week 4

The change = -370 inches

Note :

The highest change means which temperature goes the most change in water level. It can be negative also.

So, We have to take the modulus and then find the highest number.

Here, 370 > 173

So, Highest change occurs in Week 3 and Week 4 but not in Week 3 and Week 5.

7 0

3 years ago

Urgent please hell..

vovikov84 [41]

8 x (3^2/4^2)
=4.5
........................

6 0

3 years ago

Find the slope and reduce P=(-2,3) Q=(8,3)

Tanya [424]

Answer:

∆y/∆x=y2-y1/x2-x1

=3-3/(-2+3)

=0/1

=o I.e it is undefine

7 0

4 years ago

Read 2 more answers

Sec s = 1.6948

Anastaziya [24]

That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus

$\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)}
\\\\\\
thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s)
\\\\\\
\pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides}
\\\\\\
cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s$

5 0

3 years ago

Paula is located at a position of −12 feet relative to the ground. Which statement shows the distance she is from ground level?

Bess [88]

She is 12 feet below ground level

6 0

3 years ago