Answer:
2 or less packs
Step-by-step explanation:
15.99 + 1.99g < 20.00 where g is the number of packs of gum
Subtract 15.99 from each side
15.99 + 1.99g - 15.99 < 20.00-15.99
1.99g < 4.01
Divide by 1.99
1.99g/1.99 < 4.01/1.99
g < 2.015075377
Harold must buy whole packs so g =2 or less
Please plot these 3 points on graph paper. This will result in a triangle with three different vertices. Which vertex corresponds to the largest interior angle of the triangle? to the smallest? If you can answer these questions properly, then you'll pretty much have the answer to this question. Be sure to arrange the sizes of the angles in ascending (increasing) order.
<span>The two points that are most distant from (-1,0) are
exactly (1/3, 4sqrt(2)/3) and (1/3, -4sqrt(2)/3)
approximately (0.3333333, 1.885618) and (0.3333333, -1.885618)
Rewriting to express Y as a function of X, we get
4x^2 + y^2 = 4
y^2 = 4 - 4x^2
y = +/- sqrt(4 - 4x^2)
So that indicates that the range of values for X is -1 to 1.
Also the range of values for Y is from -2 to 2.
Additionally, the ellipse is centered upon the origin and is symmetrical to both the X and Y axis.
So let's just look at the positive Y values and upon finding the maximum distance, simply reflect that point across the X axis. So
y = sqrt(4-4x^2)
distance is
sqrt((x + 1)^2 + sqrt(4-4x^2)^2)
=sqrt(x^2 + 2x + 1 + 4 - 4x^2)
=sqrt(-3x^2 + 2x + 5)
And to simplify things, the maximum distance will also have the maximum squared distance, so square the equation, giving
-3x^2 + 2x + 5
Now the maximum will happen where the first derivative is equal to 0, so calculate the first derivative.
d = -3x^2 + 2x + 5
d' = -6x + 2
And set d' to 0 and solve for x, so
0 = -6x + 2
-2 = -6x
1/3 = x
So the furthest point will be where X = 1/3. Calculate those points using (1) above.
y = +/- sqrt(4 - 4x^2)
y = +/- sqrt(4 - 4(1/3)^2)
y = +/- sqrt(4 - 4(1/9))
y = +/- sqrt(4 - 4/9)
y = +/- sqrt(3 5/9)
y = +/- sqrt(32)/sqrt(9)
y = +/- 4sqrt(2)/3
y is approximately +/- 1.885618</span>
Answer:
1001.11
Step-by-step explanation:
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