John needs 20 grams of 54% acid solution for his science project. His school's science lab has bottles of 30% solution and bottles of 60% solution. Hiw much of each should john use?

The amount of 30% and 60% bottles he should join is; 4grams and 16grams respectively.

The total amount John needs is; 20 grams.

Therefore, we have;

Therefore, upon evaluation;

Therefore, The amount of 30% and 60% bottles he should join is; 4grams and 16grams respectively.

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John should use:

4 grams of the 30% solution and 16 grams of the 60% solution

Step-by-step explanation:

Let the number of grams of the 30% solution = x

Let the number of grams of the 60% solution = y

John needs 20 grams of 54% acid solution for his science project.

Hence,

x + y = 20 grams Equation 1

x = 20 - y

His school's science lab has bottles of 30% solution and bottles of 60% solution.

30% × x + 60% × y = 54% × 20

0.3x + 0.6y = 10.8Equation 2

We substitute 20 - y for x in Equation 2

0.3(20 - y) + 0.6y = 10.8

6 - 0.3y + 0.6y = 10.8

- 0.3y + 0.6y = 10.8 - 6

0.3y = 4.8

y = 4.8/3

y = 16 grams

x = 20 - y

x = 20 - 16

x = 4 grams

Therefore, John should use:

4 grams of the 30% solution and 16 grams of the 60% solution

the answer is c

step-by-step explanation:

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it