Expand and simplify: 3(x-y)+4(x+2y)

2 answers:

7 0

Expanded form: 3x - 3y + 4x + 8y Simplified form: 7x + 5y. In order to do this, you need to distribute. After that, you combine your like terms.

ad-work [718]3 years ago

3 0

The answer is 3x-3y+4x+8y

You might be interested in

a van is traveling 64 7/8 miles per hour. an airplane is traveling 5 6/7 times faster. about how fast is the airplane traveling?

ValentinkaMS [17]

Answer:

it will be 736 speed faster than

7 0

3 years ago

If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.

Sophie [7]

There's no if about it,

$f(x)=x^3+3x^2-x-3
$

has a zero $f(1)=0$ so $x-1$ is a factor. That's the special case of the Remainder Theorem; since $f(1)=0$ we'll get a remainder of zero when we divide $f(x)$ by $x-1.$

At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover $f(-1)=0$ too, so $x+1$ is a factor too by the remainder theorem. I can find the third zero as well; but let's say that's out of range for most folks.

So far we have

$x^3+3x^2-x-3 = (x-1)(x+1)(x-r)$

where $r$ is the zero we haven't guessed yet. Again we could divide $f(x)$ by $(x-1)(x+1)=x^2-1$ but just looking at the constant term we must have

$-3 = -1 (1)(-r) = r$

so

$x^3+3x^2-x-3 = (x-1)(x+1)(x+3)$

We check $f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark$

We usually talk about the zeros of a function and the roots of an equation; here we have a function $f(x)$ whose zeros are

$x=1, x=-1, x=-3$

8 0

3 years ago

Read 2 more answers

Which relation is a function of x? 1 {(1,2), (7, 6), (3, 2), (1, 0), (5, 6)}. 3. x=3y²–7

avanturin [10]

9514 1404 393

Answer:

The given graph shows a function of x.

Step-by-step explanation:

Neither 1 nor 3 is a single-valued relation for some values of x. The graph shown passes the vertical line test, so does represent a function.

The graph is a relation that is a function of x.

___