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3 years ago

Given these values f(x)=x^2 -3x-4, g(x)=-2x, h(x)-x^3 Evaluate (f(g(h(x)))

1 answer:

4 0

$\begin{cases} f(x)=x^2-3x-4\\ g(x)=-2x\\ h(x)=-x^3 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{g(~~h(x)~~)}{g(~~-x^3~~)}=-2\boxed{-x^3}\implies g(~~-x^3~~)=2x^3 \\\\[-0.35em] ~\dotfill\\\\ \underset{f(~~g(~~-x^3~~)~~)}{\stackrel{f(~~g(~~h(x)~~)~~)}{f(~~2x^3~~)}}=\boxed{2x^3}^2-3\boxed{2x^3}-4\implies f(~~2x^3~~)=2^2x^{3\cdot 2}-6x^3-4 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{f(~~g(~~h(x)~~)~~)}{f(~~2x^3~~)}=4x^6-6x^3-4~\hfill$

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A repeated-measures study uses a total of n = 10 participants to compare two treatment conditions. How many scores are measured

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A shipping container will be used to transport several 80-kilogram crates across the

Leno4ka [110]

Answers:

The inequality to solve is $80x+6300 \le 26500$
The solution to the inequality is $x \le 252.5$
Interpretation: The most crates we can load is 252 since we can only load a positive whole number of crates, and because 253 is too many.

==============================================================

Explanation:

x = number of 80-kilogram crates

x is some positive whole number

1 crate weighs 80 kilograms, so x of them will weigh 80x kilograms.

This is added on top of the 6300 kg from the other cargo already on board.

We have a total weight of 80x+6300. Let's call this T.

So T = 80x+6300

We want T to be 26500 or smaller. Otherwise, we've gone over capacity.

This must mean we want $T \le 26500$ which is the same as saying $80x+6300 \le 26500$ after replacing T with 80x+6300.

---------------------------

Let's follow PEMDAS in reverse to isolate x

$80x+6300 \le 26500\\\\80x+6300-6300 \le 26500-6300 \ \text{ ... subtract 6300 from both sides}\\\\80x \le 20200\\\\\frac{80x}{80} \le \frac{20200}{80} \ \text{ ... divide both sides by 80}\\\\x \le 252.5\\\\$

The inequality sign stays the same the entire time. It would only flip if we divided both sides by a negative number.

In the last step, we get a decimal value. However, as stated earlier, x is a positive whole number.

We cannot round to 253 because that's too high. We can check x = 253 is too high by noticing that...

80x+6300 = 80*253+6300 = 26,540

which is over the 26,500 mark by 40 kg

So we must round down to the nearest whole number and find that x = 252 is the largest x value possible. Let's check to see if we're under the weight limit

80x+6300 = 80*252+6300 = 26,460

We're under the 26,500 weight limit (with 40 kg to spare)

In short, the most crates that we can load is 252 crates in addition to the other cargo that collectively weighs 6300 kg (which may or may not consist of those 80-kilogram crates).

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