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Ede4ka [16]
3 years ago
6

On a certain day, $1 is worth 30.23 Russian rubles. Omar has $75. How many rubles will he get in exchange?

Mathematics
1 answer:
STatiana [176]3 years ago
8 0
The answer is $2267.25
30.23*75=2267.25
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Please help and explain part A B and C
Umnica [9.8K]
Part A you would just distribute your 3 to your X and your 5. After doing that you would get 3x+15+x=4x. Next you would combine like terms, meaning combine your x's together that is on the same side of your equal sign. So you would add 3x and x. When finished with that you would get, 4x+15=4x. You would then subtract your 4x on both sides of your equal sign. You then would get 15=0 which is no solution.

Part B you would distribute your 4 to your 1 and -x. After doing this your equation should then look like 4-4x=5x+8. Next you would try to get your like terms together. You would add 4x on both sides of your equal sign. Your equation should then look like 4=9x+8. Next you would subtract your 8 on both sides of the equal sign because your getting your terms together. Your equation should then look like, -4=9x. This answer would be one solution.

Part C you would combine your like terms, meaning add your 2x and x together to get your equation looking like, 3x+5=5+3x. You can tell just by looking at this equation it's going to be a infinite number of solutions. 

Hope this helps! (:
8 0
3 years ago
Use the Law of Sines to find the measure of angle J to the nearest degree.
rjkz [21]
\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-----------------------------\\\\
\cfrac{sin(J)}{9.1}=\cfrac{sin(97^o)}{11}\implies sin(J)=\cfrac{9.1\cdot sin(97^o)}{11}
\\\\\\
\textit{now taking }sin^{-1}\textit{ to both sides}
\\\\\\
sin^{-1}\left[ sin(J) \right]=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)
\\\\\\
\measuredangle J=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)
8 0
3 years ago
Oliver earns $9 per hour. Write and solve an equation to find how many hours he must work to earn $315
natka813 [3]
Y=9x

315=9x

315/9=x

35=x
7 0
3 years ago
If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+21xz+2yz+105x+10y+7z=812$, find $x+y+z$.
harina [27]

Answer:

10

Step-by-step explanation:

When we simplify we get $$z(6xy+21x+2y+7)+30xy+105x+10y=812.$$

Then we continue to factor to get: (z+5)(6xy+21x+2y+7)&=847.

We then see that we can factor (6xy+21x+2y+7) into (z+5)(3x+1)(2y+7)&=847.

we then do the prime factorization of 847, which i think is,  7*11*11. we have to find the numbers that multiply to 847 and then plug them into z+5, 3x=1,2y+7.

It has to be a positive, non-negative integer, right?

We also see that 3x+1=11 so we see that x=10/3 (which wont work).

So 3x+1=7, so x=2.

So 11 has to be in another term. It has to be in 2y+7=11 so y=2

for the last term we get z+5=11 so z=6

2+2+6=10

Hope this helps and if you want please consider giving me brainliest. :)

8 0
3 years ago
JELP HURRY !!!!!!!!!
Tems11 [23]

  You should be able to set up a proportion between the sides something like (18+x)/x=(56+21)/56

5 0
3 years ago
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