To answer this you would need to first find out what 3/8 of the 5/9 would be. 3/8 would be the amount that remains after five eights were accounted for. The question says it is 5/9 of the remaining, and 3/8 of the total would be what remains. Multiply 3/8 by 5/9 to get 5/24. This represents the part that would be green. Then I added the original 5/8 for blueand 5/24 for green. The amount is 20/24, or 5/6. This represents the part that is green or blue. 1/6 would be the other color which is red. To find the total, you would solve 1/6x =48. To solve this divide 48 by 1/6. Keep change and invert your division problem. It would be 48×6. The answer is 288 total votes. You said check your work by multiplying 5/8 by 288. You get 180. 288-180=108. Multiply 5/9 by 108 to get 60. 108-60 is 48!!!
Answer:
(27x2)1/3 1/3±27×2
Step-by-step explanation:
es la cantidad que me dio al resolverlo
Answer:
1) 1/6, 1/3, 1/2, 3/4
2) 3/10, 2/5, 7/10, 4/5
3) 1/4. 7/12. 2/3, 5/6
4) 4/15. 11/30, 2/5, 7/10
Step-by-step explanation: Either convert all in a set to a common denominator, or convert to decimal fractions.
Do recall that squaring and the *radical sign* cancel each other out... like so:(
![\sqrt{a}](https://tex.z-dn.net/?f=%20%5Csqrt%7Ba%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
= a
When you put it that way, it isn't enough :P
(
![\sqrt{a}](https://tex.z-dn.net/?f=%20%5Csqrt%7Ba%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
= a
(
![\sqrt{8x+1}](https://tex.z-dn.net/?f=%20%5Csqrt%7B8x%2B1%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
=?
so you start with
(
![\sqrt{8x+1}](https://tex.z-dn.net/?f=%20%5Csqrt%7B8x%2B1%7D%20)
)
![^{2}](https://tex.z-dn.net/?f=%20%5E%7B2%7D%20)
=
![(5)^{2}](https://tex.z-dn.net/?f=%20%285%29%5E%7B2%7D%20)
8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3
To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?
(extraneous solution is a invalid solution)
x=3 not extraneous
The inverse of this function is g(x)=9x+2. if this doesn’t work, try g(x)=9x-2 :)