Answer:
Step-by-step explanation:
To solve for y, we want to isolate y.
[multiply both sides by y]
[subtract both sides by 3y]
[factor out y]
[divide both sides by (w-3)]
Now, we know that .
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 7 − x 2 y=7-x2. what are the dime
nsions of such a rectangle with the greatest possible area?
1 answer:
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The whole field is square, and its total area is 25 acres.
It's divided up into 25 little sections.
Each little section is a square, and its area is 1 acre.
So the whole thing looks like a square tray with 25 little square blocks in it.
The blocks are in 5 rows, and there are 5 blocks in each row.
There are 16 blocks around the edge of the square tray, and 9 in the middle
that don't touch the edge.
Every block costs $1,200 to start out. But the ones around the edge are
more expensive. Anything on the edge costs an extra $150 for each side
of the little block that touches the edge.
Most of the 16 little blocks around the outside have only 1 side that touches
the edge. But the ones in the <em>corners</em> of the tray have 2 sides along the edge.
You may want to have a look at the picture I attached.
That should be plenty to put you in a position to work out the costs.
It's divided up into 25 little sections.
Each little section is a square, and its area is 1 acre.
So the whole thing looks like a square tray with 25 little square blocks in it.
The blocks are in 5 rows, and there are 5 blocks in each row.
There are 16 blocks around the edge of the square tray, and 9 in the middle
that don't touch the edge.
Every block costs $1,200 to start out. But the ones around the edge are
more expensive. Anything on the edge costs an extra $150 for each side
of the little block that touches the edge.
Most of the 16 little blocks around the outside have only 1 side that touches
the edge. But the ones in the <em>corners</em> of the tray have 2 sides along the edge.
You may want to have a look at the picture I attached.
That should be plenty to put you in a position to work out the costs.
0.99.43.55☺☺☺8%*3#3*%"&'95#4$4$59
Answer:
a) The student's overall average for the class is 87.97.
b) He would need a score above 100 to get an A, which means that he could not have scored high enough on the final exam to get an A in the class.
Step-by-step explanation:
Weighed average:
To solve this question, we find the student's weighed average, multiplying each of his grade by his weights.
Grades and weights:
Scored 82.5 on the midterm, worth 30%.
Scored 88.6 on the final exam, worth 30%.
Scored 91.6 on the homework, worth 40%.
a. On a 100-point scale, what is the student's overall average for the class?
Multiplying each grade by it's weight:
The student's overall average for the class is 87.97.
b. The student was hoping to get an A in the class, which requires an overall score of 93.5 or higher. Could he have scored high enough on the final exam to get an A in the class?
Score of x on the final class, and verify that the average could be 93.5 or higher. So
He would need a score above 100 to get an A, which means that he could not have scored high enough on the final exam to get an A in the class.