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Paha777 [63]
3 years ago
5

a store increased its prices on computers by 12%. If p represents the regular price of computers, what expression could be used

to find the new prices of computers?​
Mathematics
1 answer:
8_murik_8 [283]3 years ago
7 0

Answer:ayy but imma rapper follow me on souncloud kid ice pvt

Step-by-step explanation:

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Just calculating the differential. Step by step:

y=12 \sqrt{x} -x \frac{3}{2}-10 \\  \\ y=12x^{ \frac{1}{2} } -x \frac{3}{2}-10 \ \ \ \boxed{transform \  \sqrt{x} =x^{ \frac{1}{2} }}  \\  \\  Differentiating \\  \\ y'=12. \frac{1}{2}.x^{ \frac{1}{2}-1 } -1.x^{1-1}. \frac{3}{2}-0 \\  \\  y'=6.x^{ -\frac{1}{2} } -x^{0}. \frac{3}{2} \\  \\  y'=6. \frac{1}{x^{ \frac{1}{2} }}  -1. \frac{3}{2}  \\  \\  \boxed{y'=\frac{6}{ \sqrt{x} }  -\frac{3}{2}}

If you want to know where the tangent has no slope:

\frac{6}{ \sqrt{x} } - \frac{3}{2} =0 \\  \\  \frac{6}{ \sqrt{x} } = \frac{3}{2}  \\  \\ 3 \sqrt{x} =12 \\  \\  \sqrt{x} = \frac{12}{3}  \\  \\  \sqrt{x} =4 \\  \\   (\sqrt{x})^2 =4^2 \\  \\ x=16
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Which of the values shown are potential roots of f(x) = 3x3 â€"" 13x2 â€"" 3x 45? Select all that apply.
Verdich [7]

The potential roots of the function are, \pm1, \ \pm3, \ \pm5, \ \pm9,\  \pm15, \ \pm45,\  \pm \dfrac{1}{3},\  \pm \dfrac{5}{3}

And the accurate root is 3 it can be determined by using rules of the rational root equation.

<h2>Given that,</h2>

Function; \rm f(x) = 3x^3 - 13x^2 -3x + 45

<h3>We have to determine,</h3>

Which of the values shown are potential roots of the given equation?

<h3>According to the question,</h3>

Potential roots of the polynomial are all possible roots of f(x).

\rm f(x) = 3x^3 - 13x^2 -3x + 45

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

\rm p=\dfrac{All\  the \ positive}{Negative\  factors \ of\  45}

\rm q=\dfrac{All\  the \ positive}{Negative\  factors \ of\  3}

The factor of the term 45 are,

\pm1, \ \pm3, \ \pm5, \ \pm9,\  \pm15, \ \pm45

And The factor of 3 are,

\pm1, \ \pm3

All the possible roots are,

\dfrac{p}{q} = \pm1, \ \pm3, \ \pm5, \ \pm9,\  \pm15, \ \pm45,\  \pm \dfrac{1}{3},\  \pm \dfrac{5}{3}

Now check for all the rational roots which are possible for the given function,

\rm f(x) = 3x^3 - 13x^2 -3x + 45\\\\ f(1) = 3(1)^3 - 13(1)^2 -3(1) + 45 = 3-13-3+45 = 32\neq 0\\\\ f(-1) = 3(-1)^3 - 13(-1)^2 -3(-1) + 45 =- 3-13+3+45 = 32\neq 0\\\\ f(3) = 3(3)^3 - 13(3)^2 -3(3) + 45 = 81-117-9+45 =0\\\\ f(-3) = 3(-3)^3 - 13(-3)^2 -3(-3) + 45 = -81+117+9+45 =-144\neq 0

Therefore, x = 3 is the potential root of the given function.

Hence, The potential roots of the function are, \pm1, \ \pm3, \ \pm5, \ \pm9,\  \pm15, \ \pm45,\  \pm \dfrac{1}{3},\  \pm \dfrac{5}{3}.

For more details about Potential roots refer to the link given below.

brainly.com/question/25873992

8 0
2 years ago
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