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Neko [114]
3 years ago
10

Make r the subject of the formula t=

r-3}" alt="\frac{r}{r-3}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Gnoma [55]3 years ago
3 0

Answer:

r=\frac{3t}{t-1}

Step-by-step explanation:

So we have the formula:

t=\frac{r}{r-3}

And we want to make r the subject of the formula.

So, first, let's multiply both sides by (r-3). The right will cancel. So:

t(r-3)=r

Distribute the t:

rt-3t=r

Subtract rt from both sides. The left cancels:

-3t=r-rt

Factor out an r from the right:

-3t=r(1-t)

Divide both sides by (1-t). The right cancels:

\frac{-3t}{1-t}=r

Simplify the negatives:

\frac{3t}{t-1}=r

Flip:

r=\frac{3t}{t-1}

And we're done!

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sleet_krkn [62]
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Don't let the little 't' scare you.  It doesn't matter one bit what 't' really is.  It can be
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7 0
3 years ago
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8x^2+23=823<br><br> whats going to be the answer I got 0 am I right?
Naily [24]
1) No; you are incorrect.  "0" is NOT a solution.

Plug in "0" for "x" ;

8*0² + 23 =? 823? ; 

8*0  + 23 =? 823? ;

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_________________________________________________
2)  The answers:  x = 10, -10 ; or, write as:  x = ± 10 .
_________________________________________________

To solve:

GIven:  8x² + 23 = 823 ; 

Subtract "23" from EACH SIDE of the equation:
_____________________________________________
           8x² + 23 - 23 = 823 - 23 ; 
to get:  
           8x²  = 800 ; 

Now, divide EACH SIDE of the equation by "8" ; 

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to get:  x² = 100 ; 

        Take the "square root" of EACH SIDE of the equation; to isolate "x" on one side of the equation; and to solve for "x" ; 

          √(x²) = √(100) ;

             x  = <span>± 10 .

The answers:  x = 10, -10 ; or, write as:  x = </span><span>± 10 .
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5 0
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Answer: B and D

Step-by-step explanation:

6 0
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Step-by-step explanation:

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