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2 years ago

F(x)= x(x+3)(x+1)(x-4) has zeros at x=-3

Mathematics

1 answer:

anygoal [31]2 years ago

4 0

Answer: C) Sometimes positive; sometimes negative

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Explanation:

Pick a value between x = -1 and x = 0. Let's say we go for x = -0.5

Plug this into f(x)

f(x) = x(x+3)(x+1)(x-4)

f(-0.5) = -0.5(-0.5+3)(-0.5+1)(-0.5-4)

f(-0.5) = -0.5(2.5)(0.5)(-4.5)

f(-0.5) = 2.8125

We get a positive value.

This shows that f(x) is positive on the region of -1 < x < 0

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Now pick a value between x = 0 and x = 4. I'll use x = 1

f(x) = x(x+3)(x+1)(x-4)

f(1) = 1(1+3)(1+1)(1-4)

f(1) = 1(4)(2)(-3)

f(1) = -24

Therefore, f(x) is negative on the interval 0 < x < 4

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In short, f(x) is both positive and negative on the interval -1 < x < 4

It's positive when -1 < x < 0

And it's negative when 0 < x < 4

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3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

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3 years ago

Metal A has a density of 12.7 g/cm3.

stira [4]

Using the given information, the density of metal B is 8.58 g/cm³

<h3>Calculating density </h3>

From the question, we are to determine the density of metal B

From the given information,

55 g of metal A is combined with metal B to create an alloy with mass 90 g

∴ Mass of metal B in the alloy = 90 g - 55 g = 35 g

Now, we will determine the volume of metal A in the alloy

Using the formula,

Density = Mass / Volume

∴ Volume = Mass / Density

Volume of metal A in the alloy = 55/12.7

Volume of metal A in the alloy = 4.33 cm³

Then, we will determine the volume of the alloy

Volume of the alloy = 90/10.7

Volume of the alloy = 8.41 cm³

∴ Volume of metal B in the alloy = 8.41 cm³- 4.33 cm³

Volume of metal B in the alloy = 4.08 cm³

Now, for the density of metal B

Density = mass / volume

Density of metal B = 35 / 4.08

Density of metal B = 8.58 g/cm³

Hence, the density of metal B is 8.58 g/cm³.

Learn more on Calculating density here: brainly.com/question/13986137

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2 years ago