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4 years ago

Need help to solve the maximum

Mathematics

1 answer:

Mekhanik [1.2K]4 years ago

6 0

Answer:

B. 1

Step-by-step explanation:

Your answer for A is correct.

The time it take for the sub to hit the water will be the time between when the submarine started to get lowered and when its distance from the water was zero.

The sub started to get lowered at x = 0.

Now we need to solve for the smallest positive zero of the function.

This function factors to (x - 1)(x - 7)

From this we can tell that the submarine is a water level for time x = 1 and x = 7.

The smallest of these is 1, so that is when the submarine first hit the water and 7 is when it came back out.

The sub started getting lowered at x = 0 and hit the water at x = 1 and because x is in seconds, it took 1 second for the sub to hit the water.

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3 years ago

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Ben's living room is a rectangle measuring 10 yards by 168 inches. By how many feet does the length of the room exceed the width

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16ft
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3 years ago

A customer who obtains a loan from a bank in turn becomes?

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3 years ago

Simplify

Diano4ka-milaya [45]

Answer:

$\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}$

Step-by-step explanation:

$16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}$

$=2\left(\dfrac{1}{a^3}\right)=\dfrac{2}{a^3}$

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4 years ago

Write -3+4i in the form r(cosx+isinx)

Zanzabum

$-3+4i$ is a complex number that satisfies

$\begin{cases}r\cos x=-3\\[1ex]r\sin x=4\\[1ex]r=\sqrt{(-3)^2+4^2}\end{cases}$

The last equation immediately tells you that $r=5$ .

So you have

$\begin{cases}\cos x=-\dfrac35\\[1ex]\sin x=\dfrac45\end{cases}$

Dividing the second equation by the first, you end up with

$\dfrac{\sin x}{\cos x}=\tan x=\dfrac{\dfrac45}{-\dfrac35}=-\dfrac43$

Because the argument's cosine is negative and its sine is positive, you know that $\dfrac\pi2$ . This is important to know because it's only the case that $y=\tan x\implies \arctan y=x$ whenever $-\dfrac\pi2$ . The inverse doesn't exist otherwise.

However, you can restrict the domain of the tangent function so that an inverse can be defined. By shifting the argument of tangent by $\pi$ , we have

$\tan(x-\pi)=y\implies x=\pi+\arctan y$

All this to say

$\tan x=-\dfrac43\implies x=\pi+\arctan\left(-\dfrac43\right)\approx2.2143\text{ rad}\approx126.9^\circ$

So, $-3+4i=5(\cos126.9^\circ+i\sin126.9^\circ)$ .

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3 years ago