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Ierofanga [76]
3 years ago
13

How to find the circumference of a circle that is 42m and has diameter

Mathematics
1 answer:
Nitella [24]3 years ago
6 0

Answer:

Circumference of the circle is 131.88m

Step-by-step explanation:

we have to find the circumference of the given circle.

Given:

                     Diameter=42 m

          \\diameter=2*radius\\42 =2*radius\\radius=42/2\\radius=21 m

Circumference of a circle= 2\pi r

                          =2*\pi *21\\=2*3.14*21\\=42*3.14\\=131.88

So, the circumference of the circle is 131.88m

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aliina [53]

Answer: X = 7√2

Step-by-step explanation:

Let first Consider triangle BDC,

Cos C = adjacent/ hypothenus

Cos C = 7 / x ...... (1)

Also, let consider triangle ABC

Cos C = adjacent / hypothenus

Cos C = x / 14 ....... (2)

Since angle C is the same, equate equation 1 to 2

7/ x = x / 14

Cross multiply

X^2 = 98

Make x the subject of formula

X = sqrt (98)

X = sqrt ( 49 × 2 )

X = sqrt (49) × sqrt (2)

X = 7 sqrt(2)

X = 7√2

8 0
2 years ago
Melissa scored 120 points in each game how many points did she score in 10 games
Goshia [24]

Answer:

She scored 1200 in all ten games

5 0
3 years ago
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A spherical balloon is being inflated at a rate of 3 cubic inches per second. Determine the change in the rate of the radius.How
Novay_Z [31]

Answer:

The rate rate of change of radius is \frac{1}{48\pi} inches per second when the diameter is 12 inches.

The radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Consider the provided information.

A spherical balloon is being inflated at a rate of 3 cubic inches per second.

The volume of sphere is V=\frac{4}{3}\pi r^3

Differentiate the above formula with respect to time.

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}

Substitute the respective values in the above formula,

3=4\pi 6^2\frac{dr}{dt}

\frac{1}{48\pi}=\frac{dr}{dt}

The rate rate of change of radius is \frac{1}{48\pi} inches per second when the diameter is 12 inches.

When d=16

3=4\pi 8^2\frac{dr}{dt}

\frac{3}{256\pi}=\frac{dr}{dt}

Thus, the radius is changing more rapidly when the diameter is 12 inches.

5 0
3 years ago
Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4
dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

x_1\ =\ -2

y_1\ =\ 3

z_1\ =\ -4

Distance is measure across the line

\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}

So, we can write

\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k

=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k

=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k

=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3

=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18

=>18k-24+12k+9+10k-32\ =\ 18

=>\ k\ =\dfrac{13}{8}

So,

x\ =\ 3k-4

   =\ 3\times \dfrac{13}{8}-4

   =\ \dfrac{7}{4}

y\ =\ \dfrac{4k+3}{2}

   =\ \dfrac{4\times \dfrac{13}{8}+3}{2}

   =\ \dfrac{19}{4}

z\ =\ \dfrac{5k-16}{3}

  =\ \dfrac{5\times \dfrac{13}{8}-16}{3}

   =\ \dfrac{-21}{8}

Now, the distance of point from the plane is given by,

d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}

   =\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}

   =\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}

   =\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}

   =\ \sqrt{\dfrac{1177}{64}}

   =\ 4.28

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

            = 8.56 unit

So, the distance of a point from it's image is 8.56 units.

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astra-53 [7]

Answer:

Step-by-step explanation:

radius = d/2=22/2=11

area=pi*r*r=22/7*11*11

=380.28

6 0
3 years ago
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