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2 years ago

Serious answers only

Mathematics

2 answers:

dlinn [17]2 years ago

8 0

Answer:

192

Step-by-step explanation:

4x8=32

8x4=32 32x6=192

U have to multiply all the numbers .

timama [110]2 years ago

4 0

Answer:

Step-by-step explanation:

Alright, so we have the city that needs 800 cubic feet of salt. And then we have this truck, which will transport the salt.

We know that this truck's volume is the amount of salt it could carry in one trip, so let's solve for the truck's volume.

The volume of a rectangular prism is v=lwh, so we can multiply 4 by 8 by 6. 4 times 8 is 32, and 32 times 6 is 196. So now we know that the amount of salt this truck can carry in one trip is 196.

196/1

So we need 800. We can then do an equivalent fraction:

$\frac{196}{1}$ = $\frac{800}{x}$

196x=800(1)

196x=800

800/196 is approximately 4.08.

To find the number of trips the truck will need to make, We need to round up so that we don't fall short.

The number of trips is 5.

You might be interested in

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

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