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Feliz [49]
3 years ago
13

Both fractions in the expression of 1/3 +1/3 are examples of...

Mathematics
1 answer:
kirza4 [7]3 years ago
7 0
Equivalent fractions
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Please help me someone
Brut [27]

Answer: C

Step-by-step explanation: Consider a triangle with points A, B and C. You must create a perpendicular bisector of AB and then a perpendicular bisector of BC that crosses each other. Then you must label the center of that point D, and then using a compass tool, create a circle around the triangle.

8 0
2 years ago
Read 2 more answers
Find f(x) and g(x) so the function can be expressed as y = f(g(x)). y = Two divided by x squared. + 3
kodGreya [7K]
Hello,

g(x)=x²+3
f(x)=2/x
so f(g(x))=f(x²+3)=2/(x²+3)

Explainations:

f(y)=2/y
f(4y)=2/(4y)
f(x²)=2/x²
f(x²+3)=2/(x²+3)

4 0
3 years ago
PLEASE SOVEL THIS I WILL GIVE OUT 20 POINT
Anettt [7]

9514 1404 393

Answer:

  21.25 seconds

Step-by-step explanation:

In miles per second, the speed of the jet is ...

  (720 mi/h)/(3600 s/h) = 1/5 mi/s

The speed of the sound is the same:

  (1 mi)/(5 s) = 1/5 mi/s

Then the rate of closure between the airplane and the sound is ...

  (1/5 mi/s) +(1/5 mi/s) = 2/5 mi/s

To find the time required to close a gap of 8 1/2 miles, we can use the relation ...

  time = distance/speed

  time = (8.5 mi)/(0.4 mi/s) = 21.25 s

The pilot will hear the thunder 21.25 seconds after the flash.

5 0
3 years ago
Determine if the given differential equation is exact (e^2y-y cos xy)dx+(2xe^2 y-cos xy+2y)dy=0
rewona [7]
\dfrac\partial{\partial y}\left[e^{2y}-y\cos xy\right]=2e^{2y}-\cos xy+xy\sin xy

\dfrac\partial{\partial x}\left[2xe^{2y}-y\cos xy+2y\right]=2e^{2y}+y\sin xy

The partial derivatives are not equal, so the equation is not exact.
7 0
3 years ago
Here are the endpoints of the segments BC, FG, and JK.<br> B, −67
yulyashka [42]

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}

F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}

JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill

4 0
2 years ago
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