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mafiozo [28]
3 years ago
6

Prove that the two circles shown below are similar.

Mathematics
2 answers:
cestrela7 [59]3 years ago
6 0

Answer:

Well we need to find the area of circles B and D

We’ll use the following formula,

\pi r^2

B)

The radius of the

is circle is 4 units,

so we plug 4 into \pi r^2.

pi(4)^2

4*4 =16

16 * pi = 50.2654824574

The answer is 50 rounded to the nearest whole number.

D)

The radius is 2,

pi(2)^2

2*2 = 4

pi * 4 = 12.5663706144

Or 13 Rounded to the nearest whole number.

So now we do 50 ÷ 13 = 3.84615384615

Which is about 4 rounded to the nearest whole number.

<em>Thus,</em>

<em>The two circles are similar just that circle D is multiplied by a scale factor of 4 to get circle B</em>

jolli1 [7]3 years ago
5 0

Answer:

They have a scale factor of 2

They are both the same shape

The big circle has a radius of 4 and the small circle has a radius of 2. You divide the bigger circle by the smaller circle. 4/2=2

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Answer:

4x-10

Step-by-step explanation:

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3 years ago
Please answer it correctly and I'll mark it​
kirza4 [7]

Answer:

Step-by-step explanation:

1) a very poor question indeed. It has at least two possible solutions

c = 3x + 2                                     or     c = 3(x + 2)

when x = 2

c = 3(2) + 2 = 8                           or      c = 3(2 + 2) = 12                  

when x = 5

c = 3(5) + 2 = 17                          or       c = 3(5 + 2) = 21

2)

y = x + 2

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y = 2 + 2 = 4

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8 0
2 years ago
A bag contain 3 black balls and 2 white balls.
Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

1)

a)

Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\

b)

Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

c)

Probability of at least one black( means BB or BW or WB)

 =\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}

d)

Probability of at most one black ( means WW or WB or BW)

=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}

2)

a) Probability both black without replacement

  =\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}

b) Probability  of one black and one white

 =\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}

c) Probability of at least one black ( BB or BW or WB)

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d) Probability of at most one black ( BW or WW or WB)

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