Answer:
30
Step-by-step explanation:
If X is 5, then 6X is equal to 30. You can see to be able to get 30, you have to multiply 6 to X, which in this case is 5. So 6 times 5, you will now get the answer which is 30.
Step-by-step explanation:
.............................
Part A:
Given that a<span> bag contains 10 white golf balls and 6 striped golf balls, the ratio of white to striped golf balls is 10 : 6 = 5 : 3.
Given that </span>a<span> golfer wants to add 122 golf balls to the bag, the of balls each he should add is given by:
Therefore, the golfer shold add 76 white golf balls and 46 striped golf balls.
Part B.
You did not indicate the scale factor.
Assuming the scale factor is given by D(o, k) where k can be any number, then the cordinate of the </span><span>coordinates of the image of the quadrilateral VWXY are given by:
V'(6k, 2k), W(-2k,
4k), X(-3k, -2k), Y(3k,-5k)
That is you multiply the scale factor to the coordinates of the preimage to get the coordinates of the image.
</span>
Y = (1 + x) / (1 + x^2)
y'
= [(1 + x^2)(1) - (1 + x)(2x)] / (1 + x^2)^2
= [1 + x^2 - 2x - 2x^2] / (1 + x^2)^2
= [-x^2 - 2x + 1] / (1 + x^2)^2
y''
= [(1 + x^2)^2 * (-2x - 2) - (-x^2 - 2x + 1)(2)(1 + x^2)(2x)] / (1 + x^2)^4
= [(1 + x^2)(-2x - 2) - (4x)(-x^2 - 2x + 1)] / (1 + x^2)^3
= [(-2x - 2x^3 - 2 - 2x^2) - (-4x^3 - 8x^2 + 4x)] / (1 + x^2)^3
= [-2x - 2x^3 - 2 - 2x^2 + 4x^3 + 8x^2 - 4x] / (1 + x^2)^3
= [2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3
Setting y'' to zero, we have:
y'' = 0
[2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3 = 0
(2x^3 + 6x^2 - 6x - 2) = 0
Using trial and error, you will realise that x = 1 is a root.
This means (x - 1) is a factor.
Dividing 2x^3 + 6x^2 - 6x - 2 by x - 1 using long division, you will have 2x^2 + 8x + 2.
2x^2 + 8x + 2
= 2(x^2 + 4x) + 2
= 2(x + 2)^2 - 2(2^2) + 2
= 2(x + 2)^2 - 8 + 2
= 2(x + 2)^2 - 6
Setting 2x^2 + 8x + 2 to zero, we have:
2(x + 2)^2 - 6 = 0
2(x + 2)^2 = 6
(x + 2)^2 = 3
x + 2 = sqrt(3) or = -sqrt(3)
x = -2 + sqrt(3) or x = -2 - sqrt(3)
Note that -2 - sqrt(3) < -2 + sqrt(3) < 1
We will choose random values belonging to each interval and test them out.
-5 < -2 - sqrt(3) < -2 < -2 + sqrt(3)
f''(-5) = [2(-5)^3 + 6(-5)^2 - 6(-5) - 2] / (1 + (-5)^2)^3 = -9/2197 < 0
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0
Note that one value is positive and the other is negative.
Thus, x = -2 - sqrt(3) is an inflection point.
-2 - sqrt(3) < -2 < -2 + sqrt(3) < 0 < 1
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0
Note that one value is positive and the other is negative.
Thus, x = -2 + sqrt(3) is also an inflection point.
-2 + sqrt(3) < 0 < 1 < 2
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0
f''(2) = [2(2)^3 + 6(2)^2 - 6(2) - 2] / (1 + (2)^2)^3 = 26/125 > 0
Note that one value is positive and the other is negative.
Thus, x = 1 is an inflection point.
Hence, we have three inflection points in total.
When x = -2 - sqrt(3), we have:
y
= (1 - 2 - sqrt(3)) / (1 + (-2 - sqrt(3))^2)
= (-1 - sqrt(3)) / (1 + 4 + 4sqrt(3) + 3)
= (-1 - sqrt(3)) / (8 + 4sqrt(3))
When x = -2 + sqrt(3), we have:
y
= (1 - 2 + sqrt(3)) / (1 + (-2 + sqrt(3))^2)
= (-1 + sqrt(3)) / (1 + 4 - 4sqrt(3) + 3)
= (-1 + sqrt(3)) / (8 - 4sqrt(3))
When x = 1, we have:
y
= (1 + 1) / (1 + 1^2)
= 2 / 2
= 1
Using the slope formula, we have:
(y - 1) / (x - 1) = [[(-1 + sqrt(3)) / (8 - 4sqrt(3))] - 1] / ( -2 + sqrt(3) - 1)
(y - 1) / (x - 1) = 1/4, which is the equation of the line which the inflection points at x = 1 and x = -2 + sqrt(3) lies on.
Note that I am skipping the intermediate steps for simplifying here, but the trick is to rationalise the denominator by multiplying a conjugate on both numerator and denominator.
Now, we just need to check that the inflection point at x = -2 - sqrt(3) lies on the same line as well.
L.H.S.
= [[(-1 - sqrt(3)) / (8 + 4sqrt(3))] - 1] / (-2 - sqrt(3) - 1)
= 1/4
= R.H.S.
Once again, I am skipping simplifying steps here.
<span>Anyway, this proves all three points of inflection lies on the same straight line.</span>
