The value of the 9 is ten times the value of the 9 in 329,513. The value of the 2 is one-tenth the value of the 2 in 7,562. The
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Answer:
The least score the student can get is 71 to get an overall B average at the end of the year.
Step-by-step explanation:
Let the score she needs to get be = x
We now have to take the mean of all the five subjects <em>(including x)</em> and equate it to the least score she can have to get a B grade.
Let us take the least score the student needs to get a B grade to be 85
Mean of all 5 scores =
The least score the student can get is 71 to get an overall B average at the end of the year.
First let's solve the general case for an isosceles triangle of base 2 and height k. If the triangle is drawn so the base is on the x-axis and the apex is on the y-axis, the equation for the line containing the right side is
y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
d(area)/dw = 0 = -kw +k
w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.
If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
triangle area = (1/2)(8)(8) = 32
The maximum area of the inscribed rectangle will be half this value,
16 square units
y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
d(area)/dw = 0 = -kw +k
w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.
If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
triangle area = (1/2)(8)(8) = 32
The maximum area of the inscribed rectangle will be half this value,
16 square units