1.5 quarts of 20% fruit juice and 4.5 quarts of 100% fruit juice.
Step-by-step explanation:
Let x be the number of quarts of 100% fruit juice
Let y be the number of quarts of 20% fruit juice
The equation for amount of quarts is ;
x+y=6
x=6-y-------------------equation 1
The equation for the amount of quarts of fruit juice to get 80% fruit juice will be;
x(100%)+y(20%)=80%------however you need to make 6 quarts so divide first portion of equation by 6
{x(100%)+y(20%) }/6=80%
Multiply both parts of the equations by 6
x(100%)+y(20%)=480% --------------equation 2
Use equation 1 in 2
(6-y)(100%)+y(20%)=480%
600-100y+20y=480
600-80y=480
600-480=80y
120=80y
120/80 =y
1.5=y
Use the value of y in equation 1
x=6-y , x= 6-1.5=4.5
So you have,
1.5 quarts of 20% fruit juice and 4.5 quarts of 100% fruit juice.
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Mixtures equations;brainly.com/question/11898822
Keywords : quarts,party,bottles,punch
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Answer:
The word problem is " In a class there are 8 boys and these are 2/3rd parts of total students. How many students are there".
Step-by-step explanation:
The given expression is

We need write a word problem which is solved by the above expression.
Word problem : In a class there are 8 boys and these are 2/3rd parts of total students. How many students are there.
Solution:
Let the total number student be x.

Divide both sides by 2/3.



Therefore, the total number of students is 12.
Answer:
y=x+2
Step-by-step explanation:
To find the equation of a line or linear function, you need at least two points that lie on the line. This is to find the slope or gradient of the line if it's not given. Let's use the two points (0,2) and (-2,0). The first coordinate in each point is X and the second is the y coordinate.
GRADIENT/SLOPE=[y1-y2]/[x1-x2]
=[2-0]/[0-2]=1
The equation of a linear function is given. by Y-y1=m(X-x1)
Y-2=1(X-0)
Y=X+2
VERIFICATION OF SOLUTION
You may want to verify whether you are right by substituting values of x to see if you can obtain corresponding y values.
Example: if x=0, y=0+2=2.This confirms first point
Average annual value lost: $7,390.65First-year depreciation: $3,000.00Total depreciation: $11,125.89Total depreciation percentage: 55.63%Value of vehicle at end of ownership period: $8,874.11
see attachment for graph