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Pie
3 years ago
5

What is the relationship between the center of any circle and the point that lie on the circle?

Mathematics
2 answers:
zheka24 [161]3 years ago
8 0
Hello!

The distance from the center of a circle to any point that lies on the circle is the radius. It shows how far from the center the edge of the circle is.

I hope this helps!
Arlecino [84]3 years ago
7 0
The distance between the center and any point on the circle is the radius. That's likely what your book or teacher is asking for. 

Note: double the radius and you get the diameter
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What is the value of x?enter your answer in the box .
Katarina [22]

By pythagoras Theorem.

x² +12²=13²

=>x² +144 =169

=>x² =169-144=25

=>x=√25=5 unit.

Hence x=5 is your answer.

Hope it helps...

Regards,

Leukonov/Olegion.

4 0
3 years ago
Write the number 345000 into other forms
Tomtit [17]
345,000

300,000+40,000+5,000

<span>Three hundred forty-five thousand

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3 years ago
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Mrs. Solar is remodeling a house and is looking for a contractor to do the work. She is getting quotes from 3 different contract
andrezito [222]
#b. The equation y = 50x + 200 represents the total cost that the contractor will charge for x hours of work.
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6 0
3 years ago
Simply the complex fraction 2/3/1/3
Alexxandr [17]

Answer:

the answer is 2

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3 0
3 years ago
Bill and George go target shooting together. Both shoot at atarget at the same time. Suppose Bill hits the target withprobabilit
german

Answer: (a) \frac{2}{9}       (b) \frac{6}{41}

Step-by-step explanation:

(a) P( Bill hitting the target) = 0.7        P( Bill not hitting the target) = 0.3

    P( George hitting the target) = 0.4     P(George not hitting the target) = 0.6

Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3

                                                                                            = 0.54

Chances that George hit the target is = 0.4 x 0.3 = 0.12

So given that exactly one shot hit the target, probability that it was George's shot = \frac{0.12}{0.54} = \frac{2}{9} .

(b) The numerator in the second part would be the same as of (a) part which is 0.12.

The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.

Given that the target is hit,probability that George hit it =                                                 \frac{0.4\times 0.3}{0.7\times 0.6 +0.4\times 0.3+0.4\times 0.7}  = = \frac{6}{41}                                                                        

                                                                                           

6 0
2 years ago
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