<h3>
Answer: 127</h3>
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Explanation:
The phrasing "2/3 of 4/5 of his 2nd exam on his 3rd test" is a bit clunky in my opinion. It seems more complicated than it has to be.
The student got 80 on the second exam. 4/5 of this is (4/5)*80 = 0.8*80 = 64. Then we take 2/3 of this to get (2/3)*64 = 42.667 approximately. If we assume only whole number scores are given, then this would round to 43.
Let x be the score on the fourth exam. Since 5 points of extra credit are given, the student actually got x+5 points on this exam.
So we have these scores
- first exam = 70
- second exam = 80
- third exam = 43
- fourth exam = x+5
Adding up these scores and dividing by 4 will get us the average
(sum of scores)/(number of scores) = average
(70+80+43+x+5)/4 = 80
(x+198)/4 = 80
x+198 = 4*80
x+198 = 320
x = 320 - 198
x = 122
So the student got a score of x+5 = 122+5 = 127 on the fourth exam.
Combine like terms
assuming those 2s are exponents
-3x^2+2y^2+5xy-2y+5x^2-3y^2
move all x's together all y's together and all xy's together
-3x^2+5x^2+2y^2-3y^2+5xy
combne like terms
2x^2-y^2+5xy
3 terms
1 term is negative
90% out of 105% which would be 0.85
If you look at any graph, you'll notice that at the point where the graph crosses
the y-axis, the x-value of the graph is <em><u>zero</u></em>. Does that give you any ideas ?
Just take the equation, call 'x' zero, then solve whatever is left for 'y'.
