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m_a_m_a [10]
3 years ago
8

How to calculate markup including tax and discount ? I struggling on this.

Mathematics
1 answer:
masha68 [24]3 years ago
7 0
I would say take the total amount then I believe divide or multiply (can’t remember) by the amount of tax in ur area for example 0.06
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Find the minimum and maximum value of the function on the given interval by comparing values at the critical points and endpoint
Kitty [74]

Answer:

maximum: y = 1

minimum: y = 0.

Step-by-step explanation:

Here we have the function:

y = f(x) =  √(1 + x^2 - 2x)

we want to find the minimum and maximum in the segment [0, 1]

First, we evaluate in the endpoints, which are 0 and 1.

f(0)  =√(1 + 0^2 - 2*0) = 1

f(1) = √(1 + 1^2 - 2*1) = 0

Now let's look at the critical points (the zeros of the first derivate)

To derivate our function, we can use the chain rule:

f(x) = h(g(x))

then

f'(x) = h'(g(x))*g(x)

Here we can define:

h(x) = √x

g(x) = 1 + x^2 - 2x

Then:

f(x) = h(g(x))

f'(x)  =  1/2*( 1 + x^2 - 2x)*(2x - 2)

f'(x) = (1 + x^2 - 2x)*(x - 1)

f'(x) = x^3 - 3x^2 + x - 1

this function does not have any zero in the segment [0, 1] (you can look it in the image below)

Thus, the function does not have critical points in the segment.

Then the maximum and minimum are given by the endpoints.

The maximum is 1 (when x = 0)

the minimum is 0 (when x = 1)

7 0
3 years ago
HELP ME PLEASE! IF ANY OF YOU CAN JUST TAKE THE TIME TO HELP ME, I WOULD APPRECAITE IT. PLEASE, ASAP! Marsha created this scale
GaryK [48]
The answer is 1.33in, 2in. Any questions please just ask! Thanks!
7 0
4 years ago
Read 2 more answers
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
Chocolate candy bars cost $2.00 each. If sales tax is 9%, what is the final price of three candy bars?
Tcecarenko [31]
$6.54 is the final price of the candy bars
5 0
3 years ago
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Evaluate k (t) = 10t-19 at the specified input. <br> k(-7) = <br><br> Explain your answer.
mr_godi [17]
Plug -7 for x

10*-7 = -70 - 19 = -89

<span>k(-7) = -89</span>
8 0
3 years ago
Read 2 more answers
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