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Llana [10]
3 years ago
6

Find the area and the circumference of a circle with radius 7 cm.

Mathematics
1 answer:
aniked [119]3 years ago
3 0

Answer:

C=14pi cm

A=49pi cm^2

Step-by-step explanation:

1) Circumference of a circle is 2piR and R=7cm

C=2pi(7)

C=14pi cm

2) Area of a circle is piR^2 and R=7cm

A=pi (7)^2

A=49pi cm^2

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Formula: Y2-Y1/X2-X1
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P and a are both prime numbers. Find the values of p and q so that 6*54*p/q is a perfect cube.
daser333 [38]

6x54= 324

6x50=300

6x4=24

300+24=324

The previous nearest cube root is 216 and the next cube root would be 343.

p/q could be any prime number which there are a lot of.

im going to say p is 1 and q is 3

next you would divide 1 by 3

1/3=0.666666667

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3 0
3 years ago
Any volunteer please help
den301095 [7]

Answer:

Part A)

The equation in the point-slope form is:

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Part B)

The graph of the equation is attached below.

Step-by-step explanation:

Part A)

Given

  • The point = (-2, 11)
  • m = 4/3

The point-slope form of the line equation is

y-y_1=m\left(x-x_1\right)

Here, m is the slope and (x₁, y₁) is the point

substituting the values m = 4/3 and the point (-2, 11)  in the point-slope form of the line equation

y-y_1=m\left(x-x_1\right)

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Thus, the equation in the point-slope form is:

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Part B)

As we have determined the point-slope form which passes through the point (-2, 11) and has a slope m = 4/3

The graph of the equation is attached below.

5 0
3 years ago
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