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3 years ago

Find the average monthly income for the following: Pay Period: Biweekly Net Pay: $1,425.00

1 answer:

7 0

First add them all together = $13,137.67. Then divide that by how many addens you have (in this case 5) and that gives you $2,627.53 :)

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Can y’all help me ......

Gnoma [55]

Answer:

Honestly you can't get it wrong they are all equal to 1.

Hope this helps love

Luv Ya

5 0

3 years ago

Read 2 more answers

UIL correctly combined al like terms in the expression 6x+6-4y-3+3x-2 to write an equivalent expression what is the coefficient

-BARSIC- [3]

Answer:

The answer is 8

Step-by-step explanation: mark me brainliest plz

3 0

3 years ago

What is an algebra 2 test like? I’m taking one soon and I’m stressed out about it.

jarptica [38.1K]

well its pretty easy because you just have to remember the formula

i also in algebra 2

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3 years ago

Entomologists have discovered that a linear relationship exists between the rate of chirping of crickets of a certain species an

vekshin1

Answer:

a) $y=4x-160$

b) $y=4(102)-160=248\frac{chirps}{min}$

Step-by-step explanation:

We know that theres a linear relationship between the rate of the chirping of crickets and the air temperature.

The equation of a line is:

$y=mx+b$

So, let's name our variables

x= Temperature

y=Rate of the chirping

First of all, we need to find the slope with the two given points

$x_{1}=$ 60ºF , $y_{1} =80\frac{chirps}{min}$

$x_{2} =$ 80ºF, $y_{2} =160\frac{chirps}{min}$

By,

$m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} } =\frac{160-80}{80-60}$

$m=4$

Now, the equation between the air temperature and the number of chirps is:

$y-y_{1} =m(x-x_{1} )$

$y-80=4(x-60)$

Solving for y,

a) $y=4x-160$

b) To calculate the rate at which the crickets chirp when the temperature is 102 ºF we need to evaluate y(102)

$y=4(102)-160=248\frac{chirps}{min}$

5 0

3 years ago

Q.04: (11 points) Given the polar curve r = e θ , where 0 ≤ θ ≤ 2π. Find points on the curve in the form (r, θ) where there is a

Citrus2011 [14]

I suppose the curve is $r(\theta)=e^\theta$ .

Tangent lines to the curve have slope $\frac{dy}{dx}$ ; use the chain rule to get this in polar coordinates.

$\dfrac{dy}{dx}=\dfrac{dy}{d\theta}\dfrac{d\theta}{dx}=\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

We have

$y(\theta)=r(\theta)\sin\theta\implies\dfrac{dy}{d\theta}=\dfrac{dr}{d\theta}\sin\theta+r(\theta)\cos\theta$

$x(\theta)=r(\theta)\cos\theta\implies\dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos\theta-r(\theta)\sin\theta$

$r(\theta)=e^\theta\implies\dfrac{dr}{d\theta}=e^\theta$

$\implies\dfrac{dy}{dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}$

The tangent line is horizontal when the slope is 0, which happens wherever the numerator vanishes:

$\sin\theta+\cos\theta=0\implies\sin\theta=-\cos\theta\implies\tan\theta=-1$

$\implies\theta=\boxed{-\dfrac\pi4+n\pi}$

(where $n$ is any integer)

The tangent line is vertical when the slope is infinite or undefined, which happens when the denominator is 0:

$\cos\theta-\sin\theta=0\implies\sin\theta=\cos\theta\implies\tan\theta=1$

$\implies\theta=\boxed{\dfrac\pi4+n\pi}$

6 0

4 years ago