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valina [46]
3 years ago
11

I'm calculaing tip and tax on a total cost. Do I need to add th tax to the total, then calculate tip, or calcualte tip on the or

iginal cost?
Mathematics
1 answer:
Rainbow [258]3 years ago
4 0
I think you add the tax to the original total then get the total
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Draw the graph of the equation:<br><br> 2x-y=6
DochEvi [55]
Hope this helps!, I have to have more than 20 characters so ignore this!

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3 years ago
Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie
Dvinal [7]
For part (a), you have

\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}
x=a(x-2)+b(x+3)

If x=2, then 2=b(2-3)\implies b=-2.

If x=-3, then -3=a(-3-2)\implies a=\dfrac35.

So,

\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}

In the remainder term, the denominator x^2+x+2 can't be factorized into linear components with real coefficients, since the discriminant is negative (1-4\times1\times2=-7). However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i
\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)
\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)

Then you have

\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}
x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)

When x=-\dfrac12-\dfrac{\sqrt7}2i, you have

-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)
\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib
b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)

When x=-\dfrac12+\dfrac{\sqrt7}2i, you have

-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)
\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia
a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)

So, you could write

\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}

but that may or may not be considered acceptable by that webpage.
5 0
3 years ago
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C

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2 years ago
Question 8
tamaranim1 [39]

Answer:

Step-by-step explanation:

The number is 1/2, and 1/2 squared is 1/4.

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2 years ago
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