1. ## Multiple Choice Question

Hi
I few question i am having trouble with:
1)P and Q lie on the curve $\displaystyle y=x^3$. The x coordinates of P and Q are 2 and (2+h) respectively. The gradient of chord PQ is:
A.$\displaystyle \frac{h^3-8}{h-2}$
B.$\displaystyle 12+6h$
C.12
D.$\displaystyle \frac{(2+h)^3-8}{h}$
E.$\displaystyle 12+6h+h^2$

2)The equation $\displaystyle logx=y(log3)+1$ is equivalent to the equation:
A.$\displaystyle x=10.3^y$
B.$\displaystyle x=30^y$
C.$\displaystyle x=3^y+10$
D.$\displaystyle x=y^3+10$
E.$\displaystyle x=10y^3$

3)If the angle between the lines 2y=8x+10 and 3x-6y=22 is $\displaystyle \theta$, then $\displaystyle tan\theta$ is best approximated by:
A)1.17 B)1.40 C)2 D)0.86 E)1

P.S

2. Originally Posted by Paymemoney
Hi
I few question i am having trouble with:
1)P and Q lie on the curve $\displaystyle y=x^3$. The x coordinates of P and Q are 2 and (2+h) respectively. The gradient of chord PQ is:
A.$\displaystyle \frac{h^3-8}{h-2}$
B.$\displaystyle 12+6h$
C.12
D.$\displaystyle \frac{(2+h)^3-8}{h}$
E.$\displaystyle 12+6h+h^2$

2)The equation $\displaystyle logx=y(log3)+1$ is equivalent to the equation:
A.$\displaystyle x=10.3^y$
B.$\displaystyle x=30^y$
C.$\displaystyle x=3^y+10$
D.$\displaystyle x=y^3+10$
E.$\displaystyle x=10y^3$

3)If the angle between the lines 2y=8x+10 and 3x-6y=22 is $\displaystyle \theta$, then $\displaystyle tan\theta$ is best approximated by:
A)1.17 B)1.40 C)2 D)0.86 E)1

P.S
You should only ask one question in a thread in the future

1. P = (2,8) and Q= ([2+h], [2+h]^3)

The gradient, m, is equal to $\displaystyle \frac{\Delta y}{\Delta x} = \frac{(2+h)^3-8}{(2+h-2)}$

You don't even need to expand the cubic to see the answer

======================================

2. $\displaystyle x = 10^{ylog_{10}3+1}$. Use the laws of exponents to simplify

3. ok this is what i got for question 2

$\displaystyle x=10log3^y+1$

$\displaystyle 10log = 1$

$\displaystyle 3^y+10^1$

$\displaystyle x=3^y+10$

Answer says its $\displaystyle x=10.3^y$

4. Originally Posted by Paymemoney
ok this is what i got for question 2

$\displaystyle x=10log3^y+1$

$\displaystyle 10log = 1$ <--- where did you get this from?

$\displaystyle 3^y+10^1$

$\displaystyle x=3^y+10$

Answer says its $\displaystyle x=10.3^y$
The way I would have done it is to say that $\displaystyle c^ac^b = c^{a+b}$

$\displaystyle 10^{log3^y+1} = 10^{log3^y} \cdot 10^1 = 10 \cdot 3^y$

5. Originally Posted by Paymemoney
ok this is what i got for question 2

$\displaystyle x=10log3^y+1$
You mean $\displaystyle x= 10^{log 3^y+ 1}$ and that is equal to $\displaystyle 10^1(10^{log 3^y})= 10(3^y)$

$\displaystyle 10log = 1$

$\displaystyle 3^y+10^1$

$\displaystyle x=3^y+10$

Answer says its $\displaystyle x=10.3^y$