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2 years ago

(x 2 + 3) (x 3 + 4x)(x 2 + x - i - xi) = 0

Mathematics

1 answer:

iogann1982 [59]2 years ago

6 0

The possible values of x according to the give. equation are; 0, ±√3i, ±2i, i, -1.

<h3>What are the possible values of x?</h3>

The possible values of x in the equation given in the task content represent the zeroes of the equation and can be determined as follows;

(x²+3) (x³+4x) (x²+x-i-xi) = 0.

Hence, by further factorisation; we have;

x(x²+3) (x²+4) (x-i) (x+1) = 0.

The possible values of x are therefore;

x = 0;

x²+3 = 0; x = ±√3i

x² +4 = 0; x = ±2i

x -i = 0; x = i.

x +1 = 0; x = -1

Read more on zeros of an equation;

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How do you solve this problem?

Solnce55 [7]

Use the pythagorean theorem.

3 0

3 years ago

A backyard pool has a concrete walkway around it that is 5' wide on all sides the total area of the pool and the walkway is 950'

Soloha48 [4]

The dimensions of the pool would give a quadratic equation x² + 28x - 770 = 0

<h3>How to determine the dimension</h3>

For the pool, we have that;

Let the width = x feet

Length = (x+8) feet

The pool alongside the walkway gives;

Width = x + 5 + 5 = (x + 10) feet

Length = x + 8 + 5 + 5 = (x + 18) feet

Total area of the pool with walkway = 950 square feet

Note that formula for area is given as

Area = length * width = 950

Equate the length and width

(x+18) × (x + 10) = 950

Using the FOIL method, we have;

(x × x )+ (x × 10) + (18 × x) + (18 × 10) = 950

x² + 10x + 18x + 180 -950 = 0

collect like terms

x² + 28x - 770 = 0

Thus, the dimensions of the pool would give a quadratic equation x² + 28x - 770 = 0

Learn more about area of a rectangle here:

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3 0

1 year ago

Solve this system of equations by elimination:<br><br> 8x+9y=48<br> 12x+5y=21

Olin [163]

Answer:

Let's solve your system of equations by elimination.

8x+9y=48;12x+5y=21

Steps:

Multiply the first equation by 5,and multiply the second equation by -9.

5(8x+9y=48)

−9(12x+5y=21)

Becomes:

40x+45y=240

−108x−45y=−189

Add these equations to eliminate y:

−68x=51

Then solve−68x=51for x:

−68x /−68 = 51 /−68 (Divide both sides by -68)

x= −3 /4

Now that we've found x let's plug it back in to solve for y.

Write down an original equation:

8x+9y=48

Substitute

−3 /4

8x+9y=48:

8( −3 /4 )+9y=48

9y−6=48 (Simplify both sides of the equation)

9y−6+6=48+6 (Add 6 to both sides)

9y=54

9y /9 = 54 /9 (Divide both sides by 9) y=6

<em><u>Answer: x= −3 /4 and y=6</u></em>

Hope This Helps! Have A Nice Day!!

7 0

2 years ago

Please help!!As soon as possible!!

pychu [463]

Answer: Their equations have different y-intercept but the same slope

Step-by-step explanation:

Since, the slope of the line passes through the points (2002,p) and (2011,q) is,

$m_1 = \frac{q-p}{2011-2002} = \frac{q-p}{9}$

Similarly, the slope of the line asses through the points (2,p) and (11,q) is,

$m_2 = \frac{q-p}{11-2} = \frac{q-p}{9}$

Since, $m_1 = m_2$

Hence, both line have the same slope.

Now, the equation of the line one having slope $m_1$ and passes through the point (2002,p) is,

$y - p = \frac{q-p}{9}(x-2002)$

Put x = 0 in the above equation,

We get, $y = \frac{2000(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2000(p-q)}{9}+p)$

Also, the equation of second line having slope $m_2$ and passes through the point (2,p)

$y-p=\frac{q-p}{9}(x-2)$

Put x = 0 in the above equation,

We get, $y = \frac{2(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2(p-q)}{9}+p)$

Thus, both line have the different y-intercepts.

⇒ Third option is correct.

6 0

3 years ago

A box contain 12 balls in which 4 are white 3 are blue and 5 are red.3 balls are drawn at random from the box.find the chance th

Anastasy [175]

Answer:

3/11

Step-by-step explanation:

From the above question, we have the following information

Total number of balls = 12

Number of white balls = 4

Number of blue balls = 3

Number of red balls = 5

We solve this question using combination formula

C(n, r) = nCr = n!/r!(n - r)!

We are told that 3 balls are drawn out at random.

The chance/probability of drawing out 3 balls = 12C3 = 12!/3! × (12 - 3)! = 12!/3! × 9!

= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)

= 220 ways

The chance of selecting 3 balls at random = 220

To find the chance that all the three balls are selected,

= [Chance of selecting (white ball) × Chance of selecting(blue ball) × Chance of selecting(red balls)]/ The chance/probability of drawing out 3 balls

Chance of selecting (white ball)= 4C1

Chance of selecting(blue ball) = 3C1 Chance of selecting(red balls) = 5C1

Hence,

= [4C1 × 3C1 × 5C1]/ 220

= 60/220

= 6/22

= 3/11

The chance that all three are selected is = 3/11

4 0

3 years ago