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pochemuha
2 years ago
15

Which explanation justifies how the area of a sector of a circle is derived?

Mathematics
1 answer:
nignag [31]2 years ago
3 0
The sector of a circle is drawn by drawing two radii intercepting an arc. The area of the sector is calculated by calculating first the area of the whole circle. Then, this value is multiplied by the ratio the measure of the intercepted arc to one whole revolution. The answer to this item is the THIRD OPTION. 
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If you add Natalie's age and Fred's age, the result is 42. If you add Fred's age to 3 times
Alla [95]

Answer:

N+F=42 & F+3N=70 System Solved: Natalie is 14 and Fred is 28

Step-by-step explanation:

Using variables N=natalies age and F=Freds age

N+F=42 ----> N=42-F (take this and plug it into the other equation)

F+3N=70 so F+3(42-F)=70 [simplify] F+126-3F=70 [further simplify]

-2F=-56 [divide -56 by -2] F=28 [then plug this number into the orig. equation]

So N+28=42 or N=14.

5 0
3 years ago
The midpoint of (-4,15) and (22,3)
ahrayia [7]

Answer:

(9, 9)

Step-by-step explanation:

midpoint of (-4,15) and (22,3)

=  \bigg( \frac{ - 4 + 22}{2}, \:  \:  \frac{15 + 3}{2}  \bigg) \\  \\  =  \bigg( \frac{18}{2},  \:  \:  \frac{18}{2}  \bigg) \\  \\  =  ( 9,  \:  \:  9) \\  \\

8 0
3 years ago
PLEASE I NEED HELP! I need to find the value of X rounded to the nearest tenth
antiseptic1488 [7]

65 x 11 I have 2m and a few other people that are not interested to see if they can easily get the best of their time and will

8 0
2 years ago
What is 2x + 45?<br><br> Plz help, Wanna T.a.l.k?
Gre4nikov [31]
X= 22.5 get the x by itself by dividing by 2
8 0
2 years ago
Read 2 more answers
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0
3 years ago
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