Answer:
Five touchdowns were made in the game
Step-by-step explanation:
Here, we want to know the number of touchdowns made during the game.
We proceed as follows;
Let the number of touch downs be x
So the total points earned through touchdowns is 6 * x = 6x
The number of scores is 10 times
Let the number of extra kick be y which means that the number of conversions will be (y-1)
So the total number of score times will be ;
x + y + y-1 = 10
x + 2y = 11 •••••••••(i)
Now let’s work with points
Touch down points = 6 * x = 6x
Points from extra kick = 1 * y = y
Points from 2-point conversions = 2(y-1) = 2y - 2
So;
6x + y + 2y -2 = 37
6x + 3y = 37 + 2
6x + 3y = 39
divide through by 3
2x + y = 13 •••••••(ii)
So now solve simultaneously
From ii, y = 13 - 2x
Put this into i
x + 2(13-2x) = 11
x + 26 - 4x = 11
x -4x = 11-26
-3x = -15
x = -15/-3
x = 5
There are five touchdowns in the game
Answer:
<h3>A. -2 1/7</h3>
Step-by-step explanation:
Given the expression 5(-3/7)
On simplifying, we have:
Express as a mixed fraction
Hence the correct answer is -2 1/7
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>
A cubic polynomial function has 3 zeroes, which is why it is cubic. The student says there are 4 zeroes, which is not possible.
It is decay
the decrease percentage is 50%
