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2 years ago

Find all the zeros of x^3-5x^2-x+5

Mathematics

1 answer:

AnnyKZ [126]2 years ago

3 0

The fact that the coefficients are pairwise identical (except for sign) tells you that x^2 -1 is a factor (and x-5 is the other factor). Of course, you recognize x^2-1 as a special form (the difference of squares) that is factored as (x +1)(x -1).

The zeros are x = -1, +1, +5.

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A graphing calculator confirms this.

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LA = 5x -5 B = 3 + 13 B Solve for r and then find the measure of B:

topjm [15]

Answer:

x = 9º

∠B = 40º

Step-by-step explanation:

5x-5º = 3x+13º

[add 5 to both sides]

5x = 3x+18º

[subtract 3x from both sides]

2x = 18º

[divide each side by 2]

x = 9º

/ / / / / / / / / / / / / / / / / / /

3x+13º = 3(9º)+13º=27º+13º = 40º

6 0

2 years ago

Someone please help me

allsm [11]

Answer:

Solutions: $x = \frac{-3}{ 4} + i \sqrt{39}$ , $x = \frac{-3}{4} - i \sqrt{39}$

Step-by-step explanation:

Given the quadratic equation, 2x² + 3x + 6 = 0, where a =2, b = 3, and c = 6:

Use the quadratic equation and substitute the values for a, b, and c to solve for the solutions:

$x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}$

$x = \frac{-3 +/- \sqrt{3^{2} - 4(2)(6)} }{2(2)}$

$x = \frac{-3 +/- \sqrt{9- 48} }{4}$

$x = \frac{-3 +/- \sqrt{-39} }{4}$

$x = \frac{-3 + i \sqrt{39} }{4}$ , $x = \frac{-3 - i \sqrt{39} }{4}$

Therefore, the solutions to the given quadratic equation are:

$x = -\frac{3}{ 4} + i \sqrt{39}$ , $x = -\frac{3}{4} - i \sqrt{39}$

4 0

2 years ago

5 rows of 4 snack-size boxes of cereal in its bottom layer and is 3 layers high

leva [86]

For this problem, you're finding volume. The expression for volume is l•w•h (length times width times height).

5 0

2 years ago

During an Apollo moon landing, reflecting panels were placed on the moon. This allowed Earth-based astronomers to shoot laser be

rusak2 [61]

Answer:

The distance between the astronomers and the moon was $7.56*10^{8]$ meters.

Step-by-step explanation:

We have that the speed is the distance divided by the time, so:

$v = \frac{d}{t}$

In this problem, we have that:

The reflected laser beam was observed by the astronomers 2.52 s after the laser pulse was sent. This means that $t = 2.52$ .

If the speed of light is 3.00 times 10^8 m/s, what was the distance between the astronomers and the moon?

We have that $v = 3*10^{8}$ m/s.

We have to find d. So:

$v = \frac{d}{t}$

$d = vt$

$d = 3*10^{8}*2.52$

$7.56*10^{8]$

The distance between the astronomers and the moon was $7.56*10^{8]$ meters.

7 0

3 years ago

Messages arrive to a computer server according to a Poisson distribution with a mean rate of 11 per hour. Determine the length o

Zarrin [17]

Answer:

The length of the interval during which no messages arrive is 90 seconds long.

Step-by-step explanation:

Let X = number of messages arriving on a computer server in an hour.

The mean rate of the arrival of messages is, λ = 11/ hour.

The random variable X follows a Poisson distribution with parameter λ = 11.

The probability mass function of X is:

$P(X=x)=\frac{e^{-11}11^{x}}{x!};\ x=0,1,2,3....$

It is provided that in t hours the probability of receiving 0 messages is,

P (X = 0) = 0.76

Compute the value of t as follows:

$P(X=0)=\frac{e^{-11\times t}(11\times t)^{0}}{0!}\\0.76=e^{-11\times t}\\\ln(0.76)=-11\times t\\t=-\frac{\ln(0.76)}{11} \\=0.025\ hours\\\approx90\ seconds$

Thus, the length of the interval during which no messages arrive is 90 seconds long.

6 0

3 years ago