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aleksley [76]
1 year ago
12

Janet and Nadia each play basketball, Nadia has won twice the number of games Janet has, Is it possible for Janet to have won 10

games if the sum of the games
Nadia and Janet have won together is 24?
Mathematics
1 answer:
Lapatulllka [165]1 year ago
4 0

Answer:

It is not possible for Janet to have won 10 games.

Step-by-step explanation:

Let be "x" the number of games Janet won.

We know that Nadia has won twice the number of games Janet has and the sum of the games Nadia and Janet have won together is 24.

Then, we express this situation with this equation:

So, let's check if it is possible for Janet to have won 10 games. Substitute  into the expression:

Therefore, it is not possible for Janet to have won 10 games.

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-3^(-x)-6=-3^x+10<br><br>Solve the equation below for x by graphing plz
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I would like to solve it by using math and not graphing
if you don't want to see the math, just don't scroll down
the graphical meathod is above, first line, just read it

hmm
multiply both sides by -1
3^{-x}+6=3^x-10
multiply both sides by 3^x
3^0+6(3^x)=3^{2x}-10(3^x)
1+6(3^x)=3^{2x}-10(3^x)
minus 1 from both sides and minus 6(3^x) from both sides
0=3^{2x}-16(3^x)-1
use u subsitution
u=3^x
we can rewrite it as
0=u^2-16u-1
now factor
I mean use quadratic formula
for ax^2+bx+c=0 x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}
so for 0=u^2-16u-1, a=1, b=-16, c=-1
u=\frac{-(-16)+/-\sqrt{(-16)^2-4(1)(-1)}{2(1)}
u=8+/-\sqrt{65}
remember that u=3^x so u>0
if we have u=8+√65, it's fine, but u=8-√65 is negative and not allowed
so therfor
u=8+\sqrt{65}=3^x
8+\sqrt{65}=3^x
if you take the log base 3 of both sides you get
log_3(8+\sqrt{65})=x
if you use ln then
ln(8+\sqrt{65})=xln(3)
then
\frac{ln(8+\sqrt{65})}{ln(3)}=x

8 0
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