1. ## Check my work please

1. Using the laws of indices, remove the brackets from $\displaystyle (x^3y^7)^4$

$\displaystyle (a^mb^n)^k = a^{mk}b^{nk}$

$\displaystyle (x^3y^7)^4 = x^{34}y^{74}$

2. Transpose the formula $\displaystyle v = \sqrt{x+2y}$ for y

$\displaystyle v^2 = x+2y$

$\displaystyle \frac{v^2}{2} = x + y$

$\displaystyle \frac{v^2}{2}-x=y$

3. Convert 18 degrees to radians.

18 pi/180
= pi/10r

4. Simplify $\displaystyle \frac{x^-3}{x^-2}$
$\displaystyle x^{-3}{--2}$
$\displaystyle x^{-3}{+2}$
$\displaystyle x^-1$

5. Simplify cosAtanA
$\displaystyle tanA = \frac{SinA}{CosA}$
$\displaystyle cosAtanA = cosA\frac{sinA}{cosA} = sinA$

6. Given that f(x) = x^2+1, write an expression for f(x+2)

$\displaystyle f(x) = x^2+1$
$\displaystyle f(x+2) = (x+2)^2+1$
$\displaystyle f(x+2) = x^2+4+1$
$\displaystyle f(x+2)=x^2+5$

Thanks.

2. Originally Posted by gary223
1. Using the laws of indices, remove the brackets from $\displaystyle (x^3y^7)^4$

$\displaystyle (a^mb^n)^k = a^{mk}b^{nk}$

$\displaystyle (x^3y^7)^4 = x^{34}y^{74}$

2. Transpose the formula $\displaystyle v = \sqrt{x+2y}$ for y

$\displaystyle v^2 = x+2y$

$\displaystyle \frac{v^2}{2} = x + y$

$\displaystyle \frac{v^2}{2}-x=y$

6. Given that f(x) = x^2+1, write an expression for f(x+2)

$\displaystyle f(x) = x^2+1$
$\displaystyle f(x+2) = (x+2)^2+1$
$\displaystyle f(x+2) = x^2+4+1$
$\displaystyle f(x+2)=x^2+5$

Thanks.
Hi gary223,

you have work to do for 1), 2) and 6)

Q1 The notation means "multiply the indices"

$\displaystyle (2^2)^3=[(2)2][2(2)][2(2)]=2^{2(3)}=2^6$

$\displaystyle (a^mb^n)^k=a^{m(k)}b^{n(k)}$

$\displaystyle (x^3y^7)^4=x^{3(4)}y^{7(4)}=x^{12}y^{28}$

Q2 $\displaystyle v^2=x+2y,\ v^2-x=2y,\ y=\frac{v^2-x}{2}$

Q6 $\displaystyle (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)$

3. Hello gary223
Q6 $\displaystyle (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)$
... so $\displaystyle f(x) = x^2 +4x + 4 + 1$
$\displaystyle =x^2+4x+5$

Hello gary223... so $\displaystyle f(x) = x^2 +4x + 4 + 1$
$\displaystyle =x^2+4x+5$
Hi gary223,

you have work to do for 1), 2) and 6)

Q1 The notation means "multiply the indices"

$\displaystyle (2^2)^3=[(2)2][2(2)][2(2)]=2^{2(3)}=2^6$

$\displaystyle (a^mb^n)^k=a^{m(k)}b^{n(k)}$

$\displaystyle (x^3y^7)^4=x^{3(4)}y^{7(4)}=x^{12}y^{28}$

Q2 $\displaystyle v^2=x+2y,\ v^2-x=2y,\ y=\frac{v^2-x}{2}$

Q6 $\displaystyle (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)$

So $\displaystyle x^{12}y^{28}$

and

$\displaystyle y=\frac{v^2-x}{2}$

are the correct answers or is there more to do?

5. Hello gary223
Originally Posted by gary223
So $\displaystyle x^{12}y^{28}$
$\displaystyle y=\frac{v^2-x}{2}$