All categories

3 years ago

Factor by grouping. a^2+2ab-24b^2

1 answer:

3 0

You can rewrite that as:

a^2-4ab+6ab-24b^2 then factor 1st and 2nd pair of terms.

a(a-4b)+6b(a-4b) so you have

(a+6b)(a-4b)

You might be interested in

What is the slope of the line below? If necessary, enter your answer as a

sergejj [24]

Answer:

what

Step-by-step explanation:

4 0

3 years ago

Can someone help me plssssssssssssssssssss

Inessa05 [86]

Answer:

Both pass there all higher the 50%

5 0

3 years ago

Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v

k0ka [10]

I'll use subscript notation for brevity, i.e. $\frac{\partial f}{\partial x}=f_x$ .

By the chain rule,

$z_u=z_xx_u+z_yy_u$

$z_v=z_xx_v+z_yy_v$

$z_w=z_xx_w+z_yy_w$

We have

$z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}$

and

$\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}$

When $u=1,v=1,w=0$ , we have

$\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}$

and the partial derivatives take on values of

$\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}$

So we end up with

$\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}$

3 0

3 years ago

Martha (from problem 9) has some expenses each week. Each week she buys $9.72 worth of

Verdich [7]

if Martha is not earning money to pay it back, it would be negative. Each week, she loses -9.72 dollars. so -9.72 x 3 weeks = your problem. I'll let you take it from there and solve the rest

3 0

3 years ago

Read 2 more answers

How do you show three and three force of a number like

satela [25.4K]

Answer:

3 $\frac{3}{4}$

Hope this helps :)

4 0

3 years ago