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eduard
3 years ago
8

If the diameter of a circle is 25 inches, what is the radius of the circle?

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0
Hello there, and thank you for posting your question here on brainly.

The diameter of a circle is how long it is from the left side of the circle to the right side.

The radius of a circle is the length of either the left or the right side to the middle.

So, the radius is half of a radius.

You can divide the known diameter by 2 to get your answer.

Its diameter is 25.

Divide 25 by 2.

25 / 2 = 12.5

Its radius is 12.5 in.

That would be answer choice A.

Hope this helped!! ☺♥
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3 years ago
There are two positive numbers. four times the small number plus 3 times the big number is 41. two times the small number plus t
xxTIMURxx [149]
The small number is 2 the large number is 11.
4(2)+11(3)=41
2(2)+11=15
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hope that helps
4 0
3 years ago
I need help with this pls
shtirl [24]

Answer:5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
2. Solve each given equation and show your work. Tell whether each equation has one solution, an infinite number of solutions, o
Brut [27]

Hello!

To solve algebraic equations, we need to use SADMEP. SADMEP is an acronym used only solve for x in algebraic equations. SADMEP is expanded to be: subtraction, addition, division, multiplication, exponents, and parentheses.

(a) 4 + 2(-1) = 10 + 2 (multiply)

4 + -2 = 10 + 2 (add)

2 = 12

This equation has no solutions because <u>2 is never equal to 12</u>.

(b) 30 = 10 - (6 + 10) (simplify the parentheses)

30 = 10 + -1(16) (multiply)

30 = 10 - 16 (simplify)

30 = -6

This equation has no solutions because<u> 30 and -6 is never equal to each other</u>.

(c) 8x = 4x + 4x + 10(x - x)

8x = 4x + 4x + 10(x - x) (simplify [add and subtract])

8x = 8x + 10(0) (multiply)

8x = 8x  

This equation has an infinite number of solutions because if you <u>substitute any value into the original equation</u>, <u>both sides of the equation</u> will be <u>always equal</u>.

7 0
3 years ago
Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t&gt; 0 Given yı(t) = t y2(t) = Preview
Anestetic [448]

Let y_2(t)=tv(t). Then

{y_2}'=tv'+v

{y_2}''=tv''+2v'

and substituting these into the ODE gives

t^2(tv''+2v')+3t(tv'+v)-3tv=0

t^3v''+5t^2v'=0

tv''+5v'=0

Let u(t)=v'(t), so that u'(t)=v''(t). Then the ODE is linear in u, with

tu'+5u=0

Multiply both sides by t^4, so that the left side can be condensed as the derivative of a product:

t^5u'+5t^4u=(t^5u)'=0

Integrating both sides and solving for u(t) gives

t^5u=C\implies u=Ct^{-5}

Integrate again to solve for v(t):

v=C_1t^{-6}+C_2

and finally, solve for y_2(t) by multiplying both sides by t:

tv=y_2=C_1t^{-5}+C_2t

y_1(t)=t already accounts for the t term in this solution, so the other independent solution is y_2(t)=t^{-5}.

6 0
4 years ago
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