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3 years ago

8

Draw to show how you can use models to compare 345 and 391.

1 answer:

Trava [24]3 years ago

5 0

I have drop in a file for u to see it.

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Is a ratio a rate sometimes, always, or never true

Anna71 [15]

A simplified ratio may always be a rate.

4 0

3 years ago

Read 2 more answers

Earnings You work for 4 h on Saturday and 8 h on Sunday. You also receive a $50 bonus. You earn $164. How much did you earn per

marshall27 [118]

Hello there,

best way to do this one is set up your story line....

4h + 8h + 50 = 164$...

let's make this easier to look at and combine like terms..

12h + 50 = 164$...

Beautiful!!! Now let's try and get the h alone?...

minus 50 to both sides...

12h = 114$..

now divide both sides by 12...

H= 9.50$/hour

Looks like you made 9.50$ an hour but Hey, at least you got that bonus ;)

6 0

3 years ago

Dahasolnce [82]

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$ , b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

$\sqrt{3}\cdot \csc x - 2 = 0$

$\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0$

$\sqrt{3} - 2\cdot \sin x = 0$

$\sin x = \frac{\sqrt{3}}{2}$

$x = \sin^{-1} \frac{\sqrt{3}}{2}$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

b) The equation must be simplified first:

$\cos x + 1 = - \cos x$

$2\cdot \cos x = -1$

$\cos x = -\frac{1}{2}$

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

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

7 0

3 years ago

What is the circumference and area of a circle with a radius of 8 centimeters

Maksim231197 [3]

Answer:

Area= 64pi or 201.06

Circumference: 16pi or 50.265

Step-by-step explanation:

Area= \pi r^{2}

pi (8)^2 = 64pi or 201.06

Circumference= 2pir , 2(pi)8 = 16pi or 50.2

4 0

3 years ago

(with steps please) Find the inverse Laplace transform, f(t), of the function: 16/(s-4)^3

Stells [14]

Answer: The required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

Step-by-step explanation: We are given to find the inverse Laplace transform, f(t), of the following function :

$F(s)=\dfrac{16}{(s-4)^3}.$

We have the following Laplace formula :

$L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}}\\\\\\\Rightarrow L^{-1}\{\dfrac{1}{(s-a)^{n+1}}\}=\dfrac{t^ne^{at}}{n!}.$

Therefore, we get

$f(t)\\\\=L^{-1}\{\dfrac{16}{(s-4)^3}\}\\\\\\=16\times\dfrac{t^{3-1e^{4t}}}{(3-1)!}\\\\\\=\dfrac{16}{2}t^2e^{4t}\\\\\\=8t^2e^{4t}.$

Thus, the required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

6 0

3 years ago