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Sindrei [870]
3 years ago
13

Can someone solve these equations with the work, and explain how to do them? I'm completely lost and it's due tomorrow, plus ano

ther page, thanks! :)

Mathematics
1 answer:
trapecia [35]3 years ago
4 0
Yea what's the equations??
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Question 1 (1 point)<br> Divide using any method.<br> -3x3 + 4.x2 + 2x – 5)<br> (x - 2)
valkas [14]

Answer:Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of

0

.

x

+

1

4

x

2

-

2

x

-

5

Divide the highest order term in the dividend

4

x

2

by the highest order term in divisor

x

.

4

x

x

+

1

4

x

2

-

2

x

-

5

Multiply the new quotient term by the divisor.

4

x

x

+

1

4

x

2

-

2

x

-

5

+

4

x

2

+

4

x

The expression needs to be subtracted from the dividend, so change all the signs in

4

x

2

+

4

x

4

x

x

+

1

4

x

2

-

2

x

-

5

-

4

x

2

-

4

x

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

4

x

x

+

1

4

x

2

-

2

x

-

5

-

4

x

2

-

4

x

-

6

x

Pull the next terms from the original dividend down into the current dividend.

4

x

x

+

1

4

x

2

-

2

x

-

5

-

4

x

2

-

4

x

-

6

x

-

5

Divide the highest order term in the dividend

−

6

x

by the highest order term in divisor

x

.

4

x

-

6

x

+

1

4

x

2

-

2

x

-

5

-

4

x

2

-

4

x

-

6

x

-

5

Multiply the new quotient term by the divisor.

4

x

-

6

x

+

1

4

x

2

-

2

x

-

5

-

4

x

2

-

4

x

-

6

x

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5

-

6

x

-

6

The expression needs to be subtracted from the dividend, so change all the signs in

−

6

x

−

6

4

x

-

6

x

+

1

4

x

2

-

2

x

-

5

-

4

x

2

-

4

x

-

6

x

-

5

+

6

x

+

6

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

4

x

-

6

x

+

1

4

x

2

-

2

x

-

5

-

4

x

2

-

4

x

-

6

x

-

5

+

6

x

+

6

+

1

The final answer is the quotient plus the remainder over the divisor.

4

x

−

6

+

1

x

+

1

Step-by-step explanation:

3 0
3 years ago
What times what equals 30 but add to get 40?
musickatia [10]
Lets call those two unknown numbers a, b and write the info in the problem as equations:
a*b = 30
a + b = 40
lets solve for a in the second equation and substitute in the first:
<span>a + b = 40
</span>a = 40 - b
therefore:
<span>a*b = 30
</span>(40 - b)b = 30
40b - b^2 = 30
b^2 - 40b + 30 = 0
if we apply the general quadratic equation to solve we have:
b = (40 +- √(1600 - 120))/2
b = (40 +- √(1480<span>))/2
</span>b = (40 +- 38.47)/2
There are two solutions:
<span>b1 = (40 + 38.47)/2
</span><span>b1 = 39.24
b2 = (40 - 38.47)/2
</span>b2 = 0.765
lets use the second solution <span>b2 = 0.765, and substitute in the first equation to find a:
</span><span>a*b = 30
</span>a*0.765 = 30
a = 30/0.765
a = 39.216
so the numbers are 39.216 and 0.765
5 0
3 years ago
Read 2 more answers
1/3x-4=3/4x+1. 12(1/3x)-4=12(3/4x) + 1. 4x -4=9x + 1. -5=5x. X = -1<br><br> Solve
jek_recluse [69]

Answer:

x=-12

Step-by-step explanation:

\frac{1}{3}x - 4 = \frac{3}{4} x+ 1

To remove the denominator we take LCD

LCD of 3  and 4  is 12

In the second step we multiply the whole equation by 12 not only the fractions

12*\frac{1}{3}x - 4*12= \frac{3}{4} x*12+ 1*12

Now the denominator gets removed

1/3 * 12 = 4 , also 3/4 times 12 = 3*3 = 9

4x - 48 = 9x + 12

Subtract 9x on both sides

-5x - 48 = 12

Now add 48 on both sides

-5 x = 60

Divide both sides by -5

so x= -12

The value of x= -12

4 0
3 years ago
Work out the nth term, and work out the next term in the sequence <br> 6, 20 , 34, 48
sesenic [268]
14n-8 is how you find the answer, and the next term is 62
5 0
3 years ago
The domen range of the function y = √(x² - 4) is _____.
SpyIntel [72]

{\large\underline{\sf{Solution-}}}

Given function is

\rm \longmapsto\:y =  \sqrt{ {x}^{2}  - 4}

We know

Domain of a function is defined as set of those real values of x for which function is well defined.

So, y is defined when

\rm \longmapsto\: {x}^{2} - 4 \geqslant 0

\rm \longmapsto\: {x}^{2} -  {2}^{2}  \geqslant 0

\rm \longmapsto\: (x - 2)(x + 2)  \geqslant 0

\rm\implies \:x \leqslant  - 2 \:  \: or \:  \: x \geqslant 2

\rm\implies \:x \:  \in \: ( -  \infty ,  \: - 2] \:  \cup \: [2, \:  \infty )

Now, <u>To find the range of function </u>

We know,

Range of a function is defined as set of those real values which is obtained by assigning the values to x.

So,

\rm \longmapsto\:y =  \sqrt{ {x}^{2}  - 4}

On squaring both sides, we get

\rm \longmapsto\: {y}^{2} =  {x}^{2} - 4

\rm \longmapsto\: {y}^{2}  + 4=  {x}^{2}

\rm \longmapsto\:x =  \sqrt{ {y}^{2} + 4 }

Now, x is defined when

\rm \longmapsto\: {y}^{2} + 4 \geqslant 0 \: which \: is \: always \: true \:  \forall \: y \in \: real \: number

But,

\rm \longmapsto\:y \geqslant 0

<u>Hence, </u>

\rm\implies \:Range \: of \: function \:  = [0, \:  \infty )

So, we have

\rm\implies \:Domain \: of \: function : x \:  \in \: ( -  \infty ,  \: - 2] \:  \cup \: [2, \:  \infty )

and

\rm\implies \:Range \: of \: function \:  = [0, \:  \infty )

7 0
2 years ago
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