<u>Question 8</u>
a^2 + 7a + 12
= (a+3)(a+4)
When factorising a quadratic, the product of the two factors should equal the constant term (12), and the sum of the two factors should equal the linear term (7). To find the two factors, list out the factors of 12 (1x12, 2x6, 3x4) and identify the pair that adds up to 7 (3+4).
An alternative method if you get stuck during your exam would be to solve it algebraically using the quadratic formula and then write it in the factorised form.
a = (-7 +or- sqrt(7^2 - 4(1)(12)) / 2(1)
= (-7 +or- sqrt(1))/2
= -3 or -4
These factors are the negative of the values that would go in the brackets when written in factorised form, as when a = -3 the factor (a+3) would equal 0. (If it were positive 3 instead, then in the factorised form it would be a-3).
<u>Question 10</u>
-3(x - y)/9 + (4x - 7y)/2 - (x + y)/18
Rewrite each fraction with a common denominator so you can combine the fractions into one.
= -6(x - y)/18 + 9(4x - 7y)/18 - (x + y)/18
= (-6(x - y) + 9(4x - 7y) - (x + y)) /18
Expand the brackets and collect like terms.
= (-6x + 6y + 36x - 63y - x - y)/18
= (29x - 58y)/18
= 29/18 x - 29/9 y
Answer:
The regular price of the balls is $8
Step-by-step explanation:
The sporting goods store sales promotion is as follows;
The price of the third ball after buying two balls at regular price = $1.00
The price of the number of balls Coach John pays for the balls he bought = $136
To buy 24 balls, we have;
2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1
Therefore;
The number of balls bought at regular price = The sum of the 2s = 16 balls
The number of balls bought for $1 = 24 - 16 = 8 balls
Let x represent the regular price of the balls, we have;
16 × x + 8 = 136
16·x = 138 - 8 = 128
x = 128/16 = 8
The regular price of the balls = x = $8.
Answer:
B
Step-by-step explanation:
7.32 meters. i hope this helps
Answer:
The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.
Step-by-step explanation:
A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.
Recall that the volume for a cylinder is given by:
Substitute:
Solve for <em>h: </em>
Recall that the surface area of a cylinder is given by:
We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.
First, substitute for <em>h</em>.
Find its derivative:
Solve for its zero(s):
![\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%280%29%20%26%3D%204%5Cpi%20r%20%20-%20%5Cfrac%7B600%7D%7Br%5E2%7D%20%5C%5C%20%5C%5C%204%5Cpi%20r%20-%20%5Cfrac%7B600%7D%7Br%5E2%7D%20%26%3D%200%20%5C%5C%20%5C%5C%204%5Cpi%20r%5E3%20-%20600%20%26%3D%200%20%5C%5C%20%5C%5C%20%5Cpi%20r%5E3%20%26%3D%20150%20%5C%5C%20%5C%5C%20r%20%26%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B150%7D%7B%5Cpi%7D%7D%20%5Capprox%203.628%5Ctext%7B%20cm%7D%5Cend%7Baligned%7D)
Hence, the radius that minimizes the surface area will be about 3.628 centimeters.
Then the height will be:
![\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Cbegin%7Baligned%7D%20h%26%3D%20%5Cfrac%7B300%7D%7B%5Cpi%5Cleft%28%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B150%7D%7B%5Cpi%7D%7D%5Cright%29%5E2%7D%20%20%5C%5C%20%5C%5C%20%26%3D%20%5Cfrac%7B60%7D%7B%5Cpi%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B180%7D%7B%5Cpi%5E2%7D%7D%7D%5Capprox%207.25%206%5Ctext%7B%20cm%7D%20%20%20%5Cend%7Baligned%7D)
In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.
