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3 years ago

At what value of x does the graph of the function f(x) have a vertical asymptote?

1 answer:

3 0

It will have a vertical asymptote when the denominator approaches zero.

2x-6=0

2x=6

x=3

So the vertical asymptote is the vertical line x=3

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Eleanor can drive an average of 374 miles on one tank of gas. How many miles can she drive on 15 tanks of gas?

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Well you would just multiply 374 by 15 and that would be your answer

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A swimming pool has about 603 cubic feet of water. The pool liner has a small hole and is leaking at a rate of 1.5 cubic feet ea

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Answer:

Ruby’s function notation V(h) describes the volume of the pool after h number of hours. Hours is the independent variable and the total volume is the dependent variable.

Step-by-step explanation:

sample response

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How do you graph y=-1.5<img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20" id="TexFormula1" title=" x^{2} " alt=" x^{2} " align

lorasvet [3.4K]

$y=-1.5 x^{2} +6 \\ \\ x=-3 \Rightarrow y=-1.5*(-3)^2+6 =-1.5*9+6=-13.5+6=-7.5\\ \\x=-2 \Rightarrow y=-1.5*(-2)^2+6 =-1.5*4+6=-6+6=0\\ \\x=-1 \Rightarrow y=-1.5*(-1)^2+6 =-1.5*1+6=-1.5+6=-4.5\\ \\x=0 \Rightarrow y=-1.5*0^2+6 =6$

$x=1 \Rightarrow y=-1.5*(1)^2+6 =-1.5+6=-4.5\\ \\ x= 2 \Rightarrow y=-1.5*2^2+6 =-1.5*4+6=-6+6=0\\ \\x=3 \Rightarrow y=-1.5*3^2+6 =-1.5*9+6=-13.5+6=-7.5$

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3 years ago

jobs gross pay is $400, from which $24.80 deducted for OASDI, $5.20 for Medicare, and $45 for income tax. what is her net pay? A

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Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found: