# Math Help - Check my work please

1. ## Check my work please

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

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

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

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

$v^2 = x+2y$

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

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

3. Convert 18 degrees to radians.

18 pi/180
= pi/10r
= 0.314 radians

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

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

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

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

Thanks.

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

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

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

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

$v^2 = x+2y$

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

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

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

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

Thanks.
Hi gary223,

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

Q1 The notation means "multiply the indices"

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

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

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

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

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

3. Hello gary223
Originally Posted by Archie Meade
Q6 $(x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)$
... so $f(x) = x^2 +4x + 4 + 1$
$=x^2+4x+5$
Grandad

4. Originally Posted by Grandad
Hello gary223... so $f(x) = x^2 +4x + 4 + 1$
$=x^2+4x+5$
Grandad
Originally Posted by Archie Meade
Hi gary223,

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

Q1 The notation means "multiply the indices"

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

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

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

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

Q6 $(x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)$
Thanks Grandad, I got the same answer.

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

and

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

are the correct answers or is there more to do?

5. Hello gary223
Originally Posted by gary223
Thanks Grandad, I got the same answer.

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

and

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

are the correct answers or is there more to do?
No, they're complete.

Grandad