Answer:
61.25%
Step-by-step explanation:
How many ml of 20 percent acid should be added to pure acid
to make 70 ml of 30 percent acid?
:
let x = amt of 20% mixture
then
(70-x) = amt of pure acid
:
0.20x + (70-x) =0 .30(70)
0.20x + 70 - x = 21
= 21 - 70
-0.8x =-49
x = 61.25%
x = +61.25 ml of 20% stuff
Answer:
No solutions
Explanation:
The given system of equations is
2y = x + 9
3x - 6y = -15
To solve the system, we first need to solve the first equation for x, so
2y = x + 9
2y - 9 = x + 9 - 9
2y - 9 = x
Then, replace x = 2y - 9 on the second equation
3x - 6y = -15
3(2y - 9) - 6y = -15
3(2y) + 3(-9) - 6y = -15
6y - 27 - 6y = -15
-27 = -15
Since -27 is not equal to -15, we get that this system of equation doesn't have solutions.
Answer:
Substitute y=x-2y=x−2 into y=-0.5x+4y=−0.5x+4.
x-2=-0.5x+4x−2=−0.5x+4
2 Solve for xx in x-2=-0.5x+4x−2=−0.5x+4.
x=4x=4
3 Substitute x=4x=4 into y=x-2y=x−2.
y=2y=2
4 Therefore,
\begin{aligned}&x=4\\&y=2\end{aligned}
x=4
y=2
Step-by-step explanation:
Answer is in a photo. I couldn't attach it here, but I uploaded it to a file hosting. link below! Good Luck!
bit.
ly/3a8Nt8n
