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4 years ago

A) 40 B) 38 C) 24 D) 30

Mathematics

2 answers:

Ket [755]4 years ago

8 0

The answer would be A)40

diamong [38]4 years ago

4 0

Yes it would be (A) 40

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Let f be the function defined by f(x)=cx−5x^2/2x^2+ax+b, where a, b, and c are constants. The graph of f has a vertical asymptot

Musya8 [376]

Answer:

a) $a = 2$ and $b = -4$ , b) $c = -10$ , c) $f(-2) = -\frac{5}{3}$ , d) $y = -\frac{5}{2}$ .

Step-by-step explanation:

a) After we read the statement carefully, we find that rational-polyomic function has the following characteristics:

1) A root of the polynomial at numerator is -2. (Removable discontinuity)

2) Roots of the polynomial at denominator are 1 and -2, respectively. (Vertical asymptote and removable discontinuity.

We analyze each polynomial by factorization and direct comparison to determine the values of $a$ , $b$ and $c$ .

Denominator

i) $(x+2)\cdot (x-1) = 0$ Given

ii) $x^{2} + x-2 = 0$ Factorization

iii) $2\cdot x^{2}+2\cdot x -4 = 0$ Compatibility with multiplication/Cancellative Property/Result

After a quick comparison, we conclude that $a = 2$ and $b = -4$

b) The numerator is analyzed by applying the same approached of the previous item:

Numerator

i) $c\cdot x - 5\cdot x^{2} = 0$ Given

ii) $x \cdot (c-5\cdot x) = 0$ Distributive Property

iii) $(-5\cdot x)\cdot \left(x-\frac{c}{5}\right)=0$ Distributive and Associative Properties/ $(-a)\cdot b = -a\cdot b$ /Result

As we know, this polynomial has $x = -2$ as one of its roots and therefore, the following identity must be met:

i) $\left(x -\frac{c}{5}\right) = (x+2)$ Given

ii) $\frac{c}{5} = -2$ Compatibility with addition/Modulative property/Existence of additive inverse.

iii) $c = -10$ Definition of division/Existence of multiplicative inverse/Compatibility with multiplication/Modulative property/Result

The value of $c$ is -10.

c) We can rewrite the rational function as:

$f(x) = \frac{(-5\cdot x)\cdot \left(x+2 \right)}{2\cdot (x+2)\cdot (x-1)}$

After eliminating the removable discontinuity, the function becomes:

$f(x) = -\frac{5}{2}\cdot \left(\frac{x}{x-1}\right)$

At $x = -2$ , we find that $f(-2)$ is:

$f(-2) = -\frac{5}{2}\cdot \left[\frac{(-2)}{(-2)-1} \right]$

$f(-2) = -\frac{5}{3}$

d) The value of the horizontal asympote is equal to the limit of the rational function tending toward $\pm \infty$ . That is:

$y = \lim_{x \to \pm\infty} \frac{-10\cdot x-5\cdot x^{2}}{2\cdot x^{2}+2\cdot x -4}$ Given

$y = \lim_{x \to \infty} \left[\left(\frac{-10\cdot x-5\cdot x^{2}}{2\cdot x^{2}+2\cdot x-4}\right)\cdot 1\right]$ Modulative Property

$y = \lim_{x \to \infty} \left[\left(\frac{-10\cdot x-5\cdot x^{2}}{2\cdot x^{2}+2\cdot x-4}\right)\cdot \left(\frac{x^{2}}{x^{2}} \right)\right]$ Existence of Multiplicative Inverse/Definition of Division

$y = \lim_{x \to \pm \infty} \left(\frac{\frac{-10\cdot x-5\cdot x^{2}}{x^{2}} }{\frac{2\cdot x^{2}+2\cdot x -4}{x^{2}} } \right)$ $\frac{\frac{x}{y} }{\frac{w}{z} } = \frac{x\cdot z}{y\cdot w}$

$y = \lim_{x \to \pm \infty} \left(\frac{-\frac{10}{x}-5 }{2+\frac{2}{x}-\frac{4}{x^{2}} } \right)$ $\frac{x}{y} + \frac{z}{y} = \frac{x+z}{y}$ / $x^{m}\cdot x^{n} = x^{m+n}$

$y = -\frac{5}{2}$ Limit properties/ $\lim_{x \to \pm \infty} \frac{1}{x^{n}} = 0$ , for $n \geq 1$

The horizontal asymptote to the graph of f is $y = -\frac{5}{2}$ .

4 0

4 years ago

What is the slope of the line that passes through the pair of points (3/2,-2) and (-3,7/3)?

lilavasa [31]

Slope = (7/3 + 2)/(-3 - 3/2)
slope = 13/3 / -9/2
slope = 13/3 x (-2/9)
slope = -26/27

3 0

4 years ago

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple

Veseljchak [2.6K]

Answer:

The 95% confidence interval is (-0.2451, 06912)

Step-by-step explanation:

From the question, we have;

The number of small cars in the sample of small cars, n₁ = 12

The number of small cars that were totaled, x = 8

The number of large cars in the sample of small cars, n₂ = 15

The number of large cars that were totaled, y = 5

Therefore, the proportion of small cars that were totaled, pX = x/n₁

∴ pX = 8/12 = 2/3

The proportion of large cars that were totaled, pY = y/n₁

∴ pY = 5/15 = 1/3

The 95% confidence interval for the difference pX - pY is given as follows;

$pX-pY\pm z^{*}\sqrt{\dfrac{pX\left (1-pX \right )}{n_{1}}+\dfrac{pY\left (1-pY \right )}{n_{2}}}$

$\dfrac{2}{3} -\dfrac{1}{3} \pm 1.96 \times \sqrt{\dfrac{\dfrac{2}{3} \times \left (1-\dfrac{2}{3} \right )}{12}+\dfrac{\dfrac{1}{3} \times \left (1-\dfrac{1}{3} \right )}{15}}$

Therefore, we have;

$\therefore 95\% \ CI = \dfrac{1}{3} \pm 0.3578454$

The 95% confidence interval, CI = (-0.2451, 06912)

6 0

3 years ago

What is the answer to <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bv%7D%7B8%7D%20%20%3D%2010" id="TexFormula1" title=" \fr

Shkiper50 [21]

Answer:

Step-by-step explanation:

$\frac{v}{8} =10\\v=10*8\\v=80$

7 0

3 years ago

Read 2 more answers

Josh had $10 more than Carly, so he gave Carly half of his money. Then Carly had more money than josh, so she gave Josh $4. Then

DedPeter [7]

each person starts with $8.

<u>Step-by-step explanation:</u>

Here we have , Josh had $10 more than Carly, so he gave Carly half of his money. Then Carly had more money than josh, so she gave Josh $4. Then they both had $13. We need to find How much money did each person start . Let's find out:

Suppose Carly had $x , Josh had $10 more than Carly i.e. $(x+10), so he gave Carly half of his money i.e.

⇒ $\frac{x+10}{2}$ ,Now , Josh have $ $\frac{x+10}{2}$ & Carly have $x + \frac{x+10}{2}$ .