1 Writing and 3 Maths Questions

Test Preparation
1

<p>Hi,</p>

<p>There's just a few questions in the BB I still don't understand - if anyone could help with all (or any) of them it would be hugely appreciated!!</p>

<p>Page 835: Question 15-</p>

<p>If x^2-y^2=10 and x + y = 5, what is the value of x-y?</p>

<p>I tried to do simultaneous equations but didn't get anywhere...</p>

<p>The answer is 2.</p>

<p>Page 906: Q 12-</p>

<p>If y is directly proportional to x^2 and y=1/8 when x=1/2, what is the positive value of x when y=9/2?</p>

<p>The answer is 3.</p>

<p>Page 906: Q 14-</p>

<p>Let the function h be defined by h(t)=2(t^3-3). When h(t)=-60, what is the value of 2-3t?</p>

<p>The answer is 11.</p>

<p>And finally one writing question.</p>

<p>Page 839: Q 29-</p>

<p>Today a medical doctor must often make a choice between engaging in private practice or (error) engaging in research.</p>

<p>The error is with or, but I don't know why...</p>

<p>Thanks in advance.</p>

2

<p>for the first question you have a typo, it’s not what is the value of xy, its what is the value of x-y</p>

<p>If x^2-y^2=10 and x + y = 5, what is the value of x-y?</p>

<p>So, back to the question, x^2-y^2=10 can be factorized to (x+y)(x-y)=10 </p>

<p>Since (x+y) = 5, substitute it in the expression so it reads 5(x-y)=10
Divede both sides of the eqaution by 5 and you get x-y = 2</p>

3

<p>Oh, yeah, it is.</p>

<p>Thanks, I’ve just corrected it.
I still have no idea how to do it unfortunately…</p>

4

<p>Question 15 is 5…</p>

<p>5squared + 0squared = 10…5+0= 5…so. 5-0= 5</p>

<p>Sent from my SPH-M920 using CC App</p>

5

<p>Whoops ignore me I looked @ the question wrong…lol</p>

<p>Sent from my SPH-M920 using CC App</p>

6

<p>as for question 14,</p>

<p>-60 = 2(t^3 -3), the goal is to find the value of t first so we can find the value of 2-3t.</p>

<p>first, divide both sides of the equation by 2 so it reads -30= t^3 - 3
Add 3 to both sides to get -30+3=t^3 = -27=t^3
now find the cube root of both sides to get t = -3</p>

<p>now substitute the value of t in the equation 2-3t
2-3(-3) = 2+9 = 11</p>

<p>Hope this helps! :)</p>

<p>I know how to get the answer for 12, but i don’t know how to explain it easily.</p>

7

<p>Wow, I have no idea how I missed that… Seems so obvious now you did it!</p>

8

<p>I believe the correct answer to the writing is “and” in place of “or”.
The “or” insinuates more than two items being compared. I may be wrong though.</p>

9

<p>Thanks very much - I think you’re right.
Does anyone have an explanation for question 12?</p>

10

<p>No-one got anything for q. 12?</p>

11

<p>I’ll try my best to explain it…</p>

<p>y is directly proportional to x^2, simply means that y is a constant multiple of x^2, this is represented by the equation y=kx^2. where k is the constant.</p>

<p>To determine k we have to substitute the values of y and x n the equation.
y=kx^2
1/8 = k (1/2)^2<br>
1/8 = k(1/4)
k= (1/8) / (1/4)
k = 1/2</p>

<p>So we know now that k = 1/2</p>

<p>the next part of the question ask us to determine the value of x, when y = 9/2. So we use our equation</p>

<p>y=kx^2
9/2= (1/2) x^2
(9/2) / (1/2) = x^2
9= x^2
9 is the value of x^2, but we want the value of x so to find that we take the sqaure root of both sides of the equal sign to get x
3=x</p>

12

<p>For question 15, you need to do a system of equations. Get x+y=5 to equal y=X-5. Substitute y into x^2-y^2=10. So you get x^2-(5-x)=10. Solve, you get x = 3.5 plug that into x+y because its an easier equation to solve then you get y=-1.5 add them, you get 2 :)</p>

13

<p>Oops, I meant y=5-x, not y=x-5.</p>