**30 POINTS**

Simplify. Assume that no denominator is equal to zero. ([3^2]^3g^3h^4)^2

mariarad [96]

Answer:

531,441·g^6·h^8

Step-by-step explanation:

The operative rule of exponents is ...

(a^b)^c = a^(b·c)

Working from the inside out, according to the order of operations, we get ...

= (9^3·g^3·h^4)^2

= 729^2·g^(3·2)·h^(4·2)

= 531,441·g^6·h^8

7 0

3 years ago

The inequality of the graph

fiasKO [112]

Answer:

$\displaystyle y > 4x - 2$

Step-by-step explanation:

Starting from the y-intercept of $\displaystyle [0, -2],$ you do $\displaystyle \frac{rise}{run}$ by either moving four blocks south over one block west or four blocks north over one block east [west and south are negatives]. Next, we have to determine the types of inequality symbols that are suitable for this graph, which will be less than and greater than since this is a dashed line graph. We then use the zero-interval test [test point (0, 0)] to ensure whether we shade the opposite portion [portion that does not contain the origin] or the portion that DOES contain the origin. At this step, we must verify the inequalities as false or true:

Greater than

$\displaystyle 0 > 4[0] - 2 → 0 > -2$ ☑

Less than

$\displaystyle 0 < 4[0] - 2 → 0 ≮ -2$

This graph is shaded in the portion of the origin, so you would choose the greater than inequality symbol to get this inequality:

$\displaystyle y > 4x - 2$

I am joyous to assist you anytime.

3 0

3 years ago

X=-3 which is a positive value? A x B x^2 C x^3 D x+3

Gemiola [76]

Answer:

B. X squared is -3 times -3, which is equal to 9.

Step-by-step explanation:

5 0

3 years ago

PLSSS HELP ME!!!!!!!!!!! I WILL MARK U!!!!!

max2010maxim [7]

Answer:

a2

Step-by-step explanation:

you're using the pythagorean theorem

8 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

2 years ago

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