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2 years ago

Raise c to the 3rd power, then triple the result

Mathematics

1 answer:

777dan777 [17]2 years ago

5 0

Step-by-step explanation:

Raise c to the 3rd power:

${c}^{3}$

Triple the amount (Multiply by 3):

$3( {c}^{3} )$

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How can I solve this?

Mnenie [13.5K]

X-1/3 = y-2/4
4(x-1) = 3(y-2)
4x-4 = 3y-6
4x-3y = -2 --->1
4x +3y =8 ---->2
2-1 ; 6y = 10
y = 10/6 or 5/3
And 4x + 3(10/6) = 8
4x =3
X = 3/4

6 0

3 years ago

The present above is a 10 in by 10 in by 10 in cube. How many square inches of wrapping paper do you need to wrap the box?

Mashutka [201]

Answer:

600 in^2.

Step-by-step explanation:

There are 6 faces on a cube so the area we need is:

6 * 10 * 10

= 600 in^2.

5 0

2 years ago

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$