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3 years ago

Need help on solving this ?

Mathematics

1 answer:

katovenus [111]3 years ago

4 0

For the first question, simply plug in the value of 3 into the polynomial question. The final answer is your volume of the hot water tub.

For the second question, you can figure out that x = -5 is not a real root because it does not touch or cross the x axis.

If it does it would be an solution.

Read more about circles at:

brainly.com/question/11833983

#SPJ1

5 0

2 years ago

Find the y-intercept of the parabola whose equation is f(x)=-2x^2-8x+5

tatyana61 [14]

Answer:

i'm pretty sure the y-intercept would be 5

6 0

3 years ago

What is the z-score of a value that is 2.08 standard deviations greater than the mean ?

ser-zykov [4K]

Need help on the same question

7 0

4 years ago

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

Use the Factor Theorem and synthetic division to show x + 5 is a factor of f(x) = 2x3 + 7x2 − 14x + 5

olga2289 [7]

Answer:

Factor theorem: $f(-5) = 0$ .

Synthetic division: $f(x) = (x + 5)\, (2\, x^2 -3\, x + 1)$ .

Step-by-step explanation:

<h3>Factor Theorem</h3>

By the factor theorem, a monomial of the form $(x - a)$ (where $a$ is a constant) is a factor of polynomial $f(x)$ if and only if $f(a) = 0$ .

In this question, the monomial is $(x + 5)$ , which is equivalently $(x - (-5))$ . $a = -5$ .

$\begin{aligned}& f(-5) \\ &= 2 \times (-5)^{3} + 7 \times (-5)^{2}- 14\times (-5) + 5\\ &= -250 + 175 + 70 + 5 \\ &= 0\end{aligned}$ .

Hence, by the factor theorem, $(x + 5)$ , which is equivalent to $(x - (-5))$ , is a factor of $f(x)$ .

<h3>Synthetic Division</h3>

$\begin{aligned}& f(x) \\ &= 2\, x^{3} + 7\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} - 3\, x^{2} - 14\, x + 5 \\ &= \underbrace{(x + 5) \, (2\, x^2)}_{2\, x^{3} + 10\, x^{2}} + \underbrace{(x + 5)\, (-3\, x)}_{-3\, x^{2} - 15\, x} + (x + 5) \\ &= (x + 5)\, (2\, x^{2} - 3\, x + 1)\end{aligned}$ .

The remainder is $0$ when dividing $f(x)$ by $(x + 5)$ . Hence, $(x + 5)\!$ is a factor of $f(x)\!$ .

6 0

3 years ago