You should have drawn1 - x-axis and y-axis in light pencil.2 - graphed a down-facing parabola with the top of the frown on the y-axis at y = 2. It should be crossing the x-axis at ±√2. This should be in dark pencil or another color.3 - In dark pencil or a completely new color, draw a rectangle with one of the horizontal sides sitting on top of the x-axis and the other horizontal side touching the parabola at each of the top corners of the rectangle. The rectangle will have half of its base in the positive x-axis and the other half on the negative x-axis. It should be split right down the middle by the y-axis. So each half of the base we will say is "x" units long. So the whole base is 2x units long (the x units to the right of the y-axis, and the x units to the left of the y-axis) I so wish I could draw you this picture... In the vertical direction, both vertical edges are the same length and we will call that y. The area that we want to maximize has a width 2x long, and a height of y tall. So A = 2xy This is the equation we want to maximize (take derivative and set it = 0), we call it the "primary equation", but we need it in one variable. This is where the "secondary equation" comes in. We need to find a way to change the area formula to all x's or all y's. Since it is constrained to having its height limited by the parabola, we could use the fact that y=2 - x2 to make the area formula in only x's. Substitute in place of the "y", "2 - x2" into the area formula. A = 2xy = 2x(2 - x2) then simplify A = 4x - 2x3 NOW you are ready to take the deriv and set it = 0 dA/dx = 4 - 6x2 0 = 4 - 6x2 6x2 = 4 x2 = 4/6 or 2/3 So x = ±√(2/3) Width remember was 2x. So the width is 2[√(2/3)]Height is y which is 2 - x2 = 2 - 2/3 =4/3
by the use of elimination method
make all coefficients of subject to be eliminated similar..by multiplying the coefficients with one another
for eqn(i)
5(-10y+9x=-9)
-50y+45x=-45
for eqn(ii)
9(10y+5x=-5)
90y+45x=-45
-50y+45x=-45
90y+45x=-45
...subtract each set from the other...
we get
-140y+0=0
y=0
from eqn(i)
10y+5x=-5
0+5x=-5
x= -1
Answer:
Step-by-step explanation:
Corresponding scores before and after taking the course form matched pairs.
The data for the test are the differences between the scores before and after taking the course.
μd = scores before taking the course minus scores before taking the course.
a) For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
b) We would assume a significance level of 0.05. The P-value from the test is 0.65. The p value is high. It increases the possibility of accepting the null hypothesis.
Since alpha, 0.05 < than the p value, 0.65, then we would fail to reject the null hypothesis. Therefore, it does not provide enough evidence that scores after the course are greater than the scores before the course.
c) The mean difference for the sample scores is greater than or equal to zero
Answer: 12/4 or 3
Step-by-step explanation: Switch the numerator and denominator. This should leave you with the answer above.
Using the formula of P(1 + r)^n = x where p represents the initial value, r represents the rate and n represents the number of years and x is our final output. We want to find P so we have to make it the subject of the equation.
1 + 0.04 = 1.04
1.04^18 = 2.025816515
Then divide the total amount by this to get 185,110.5454
Therefore the answer is $185,110.55
Hope this helps! :)
