He will mix some 20% solution and some 70% solution to make 60% solution.
Let x = number of liters of 20% solution.
Let y = number of liters of 70% solution.
He wants to make 50 liters of 60% solution, so
x + y = 50 First Equation
The amount of acid in x amount of 20% solution is 20% of x, or 0.2x
The amount of acid in y amount of 70% solution is 70% of y, or 0.7y
The amount of acid in 50 liters of 60% solution is 60% of 50 liters, or 0.6 * 50 = 30
Now we add the amounts of acid.
0.2x + 0.7y = 30 Second Equation
x + y = 50
0.2x + 0.7y = 30
-0.2x - 0.2y = -10
0.2x + 0.7y = 30
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0.5y = 20
y = 40
x + y = 50
x + 40 = 50
x = 10
Answer: He needs 10 liters of 20% solution and 40 liters of 70% solution.
Ok well you would use pathagorean there so A2+B2=C2
Answer:
Image wont load can you explain the problem to me through comments pls?
Step-by-step explanation:
Answer:
a reflection across y-axis then translation of 1 unit right and 2 units up.
a clockwise rotation of 180° about the origin then a translation of 1 unit right and 3 units up
a reflection across y-axis then translation of 1 unit right and 1 unit down
a reflection across y = x then a positive rotation of 270° about the origin
Step-by-step explanation:
Answer:
ASA and AAS
Step-by-step explanation:
We do not know if these are right triangles; therefore we cannot use HL to prove congruence.
We do not have 2 or 3 sides marked congruent; therefore we cannot use SSS or SAS to prove congruence.
We are given that EF is parallel to HJ. This makes EJ a transversal. This also means that ∠HJG and ∠GEF are alternate interior angles and are therefore congruent. We also know that ∠EGF and ∠HGJ are vertical angles and are congruent. This gives us two angles and a non-included side, which is the AAS congruence theorem.
Since EF and HJ are parallel and EJ is a transversal, ∠JHG and ∠EFG are alternate interior angles and are congruent. Again we have that ∠EGF and ∠HGJ are vertical angles and are congruent; this gives us two angles and an included side, which is the ASA congruence theorem.
