$The\ equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\where\ (a;\ b)\ -the\ coordinates\ of\ the\ center;\ r-the\ radius\\-----------------------------\\(-1;-2)-the\ center\ of\ the\ circle\\r=5\\\\The\ equation:(x+1)^2+(y+2)^2=5^5\\\\Put\ x=2\ to\ the\ equation:\\\\(2+1)^2+(y+2)^2=25\\3^2+(y+2)^2=25\\9+(y+2)^2=25\\(y+2)^2=25-9\\(y+2)^2=16\iff y+1=\pm\sqrt{16}\\\\y+1=-4\ or\ y+1=4\ \ \ \ \ |subtract\ 1\ from\ both \sides\\y=-5\ or\ y=3\\\\Answer:(2;-5)\ and\ (2;\ 3).$

5 0

3 years ago

One uranium atom has a mass of 3.95 x 10^-22 grams.

creativ13 [48]

Remember
(ab)/(cd)=(a/c)(b/d)
and
$\frac{x^n}{x^m}=x^{n-m}$

1kg=1000g=1*10^3g
hmm, I estimate, 4 of them

x=number of attoms
mass of 1 times x=1*10^3
3.95*10^-22 times x=1*10^3
divide both sides by 3.95*10^-22
x= $\frac{1*10^3}{3.95*10^{-22}}= (\frac{1}{3.95} )( \frac{10^{3}}{10^{-22}} )$ = $0.25316*10^{25}$ = $2.5316*10^{24}$
a million has 9 zeroes
so 24 zeroes is alot that is a <span>septillion, so there are 2.5316 septillion uranium atoms in 1kg

a. 4
b. underestimate

</span>

5 0

3 years ago

Desperate Please Hurry<br> Question down below

VladimirAG [237]

Answer:

See below

Step-by-step explanation:

I'm guessing you want to fill in the blanks of the sequence?

We can represent this as an equation:

31-3x=18

-3x=13

x=-13/3

Second term: 31 - (13/3) = 26 2/3

Third term: 26 2/3 - (13/3) = 22 1/3

4 0

2 years ago

What is the range of the data in the table?

IgorLugansk [536]

Answer:

<h2>range = maximum value - minimum value.</h2>

5 0

3 years ago

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