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4 years ago

11

How can we show 10 minus 6?

1 answer:

neonofarm [45]4 years ago

8 0

10-6=4
You cannot use 6-10 because that would give you a negative answer.

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What is the mean of the data set?

Katarina [22]

Answer:

The answer is 26. Add them all together then divide by the amount of numbers.

Step-by-step explanation:

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3 years ago

If s(x) = 2x2 + 3x - 4, and t(x) = x + 4 then s(x) · t(x) =

Nookie1986 [14]

You simply multiply the two and distribute to solve. I am starting with t(x) because I think it is easier.

(x+4)*(2x^2+3x-4)
2x^3+3x^2-4x+8x^2+12x-16 (distribute)
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3 years ago

Read 2 more answers

What is the solution to the equation g(x)=3?

aksik [14]

Answer:

3 ≤ x < 5

Step-by-step explanation:

The top horizontal line segment on the graph corresponds to g(x) = 3. That segment extends over the interval from 3 to 5, including 3, but not including 5. So ...

g(x) = 3 for 3 ≤ x < 5 . . . . matches the last choice

3 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

So

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago

ra1l [238]

Answer:

True

Step-by-step explanation:

$x^{2} -6x=9$

subtract $9$ from both sides

$x^{2} -6x-9$

<em>plz mark me brainliest.</em> ;)

5 0

3 years ago