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Anestetic [448]
3 years ago
9

A+c=r+d solve for a, answer please :(

Mathematics
2 answers:
RoseWind [281]3 years ago
7 0

Answer:

a = r + d - c

Step-by-step explanation:

All we do is transpose c to the other side to isolate a.

Elanso [62]3 years ago
5 0

Answer:

a = r + d - c

Step-by-step explanation:

a + c = r + d

You want a alone on the left side. c is being added to a.

To get rid of the c, you must subtract c from the left side, but the rule of equations is that you must do the same operation to both sides of the equation. We subtract c from both sides of the equation.

a + c - c = r + d - c

c - c = 0, so the c is eliminated from the left side.

We now have

a + 0 = r + d - c

a = r + d - c

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trapecia [35]

Express the left hand side as

sin3theta=sin(theta+2theta)

now expand the right side of this equation using color(blue)"Addition formula"

color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))

rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))

The right hand side is expressed only in terms of sintheta's

so we use cos2theta=1-2sin^2theta........(1)

color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)

Replace cos2theta" and " sin2theta by the expansions (1) and (2)
into (A)

sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)

and expanding brackets gives.

sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))

Replace cos^2theta=1-sin^2theta" into (B)"

rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)

and expanding 2nd bracket gives.

sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta

Finally, collecting like terms.

sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"
3 0
3 years ago
The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units
ad-work [718]
Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&A&(~ 2 &,& 8~) 
%  (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
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\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&A&(~ 2 &,& 8~) 
%  (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
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AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}

so, add AC + AB + CB, and that's the perimeter of the triangle.

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3 years ago
Read 2 more answers
I really need someone to help me out
elixir [45]

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3 0
3 years ago
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