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Anastasy [175]
3 years ago
13

Hi, im new i need help

Mathematics
2 answers:
Studentka2010 [4]3 years ago
5 0

Answer:

with wat

Step-by-step explanation:

jasenka [17]3 years ago
4 0

Answer:

what do u need help with! :)

Step-by-step explanation:

You might be interested in
Negitive 8 over 2 divided by 6 over negitive 3 = ??????
solmaris [256]

when you didve fractions flip the second one over and multiply

 so -8/2 divided by 6/-3 will become

-8/2 times -3/6

then multiple straight across so -8 x -3 = 24

over 6 time 2 = 12

 so you get 24/12 which equals 2

5 0
2 years ago
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I really need it to be sold in imaginary numbers
Yuliya22 [10]
Solving a 5th grade polynomial

We want to find the answer of the following polynomial:

x^5+3x^4+3x^3+19x^2-54x-72=0

We can see that the last term is -72

We want to find all the possible numbers that can divide it. Those are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

We want to factor this polynomial in order to find all the possible x-values. In order to factor it we will have to find some binomials that can divide it using the set of divisors of -72.

We know that if

(x - z) is a divisor of this polynomial then z might be a divisor of the last term -72.

We will verify which is a divisor using synthetic division. If it is a divisor then we can factor using it:

Let's begin with

(x-z) = (x - 1)

We want to divide

\frac{(x^5+3x^4+3x^3+19x^2-54x-72)}{x-1}

Using synthetic division we have that if the remainder is 0 it will be a factor

We can find the remainder by replacing x = z in the polynomial, when it is divided by (x - z). It is to say, that if we want to know if (x -1) is a factor of the polynomial we just need to replace x by 1, and see the result:

If the result is 0 it is a factor

If it is different to 0 it is not a factor

Replacing x = 1

If we replace x = 1, we will have that:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ 1^5+3\cdot1^4+3\cdot1^3+19\cdot1^2-54\cdot1-72 \\ =1+3+3+19-54-72 \\ =-100 \end{gathered}

Then the remainder is not 0, then (x - 1) is not a factor.

Similarly we are going to apply this until we find factors:

(x - z) = (x + 1)

We replace x by -1:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ (-1)^5+3\cdot(-1)^4+3\cdot(-1)^3+19\cdot(-1)^2-54\cdot(-1)-72 \\ =-1+3-3+19+54-72 \\ =0 \end{gathered}

Then, (x + 1) is a factor.

Using synthetic division we have that:

Then:

x^5+3x^4+3x^3+19x^2-54x-72=(x+1)(x^4+2x^3+x^2+18x-72)

Now, we want to factor the 4th grade polynomial.

Let's remember our possibilities:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

Since we verified ±1, let's try with ±2 as we did before.

(x - z) = (x - 2)

We want to divide:

\frac{x^4+2x^3+x^2+18x-72}{x-2}

We replace x by z = 2:

\begin{gathered} x^4+2x^3+x^2+18x-72 \\ \downarrow \\ 2^4+2\cdot2^3+2^2+18\cdot2-72 \\ =16+16+4+36-72 \\ =0 \end{gathered}

Then (x - 2) is a factor. Let's do the synthetic division:

Then,

x^4+2x^3+x^2+18x-72=(x-2)(x^3+4x^2+9x+36)

Then, our original polynomial is:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =\mleft(x+1\mright)\mleft(x^4+2x^3+x^2+18x-72\mright) \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \end{gathered}

Now, let's prove if (x +2) is a factor, using the new 3th grade polynomial.

(x - z) = (x + 2)

We replace x by z = -2:

\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-2)^3+4(-2)^2+9(-2)+36 \\ =-8+16-18+36 \\ =26 \end{gathered}

Since the remainder is not 0, (x +2) is not a factor.

All the possible cases are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

let's prove with +4

(x - z) = (x + 4)

We want to divide:

\frac{x^3+4x^2+9x+36}{x+4}

Let's replace x by z = -4 in order to find the remainder:

\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-4)^3+4(-4)^2+9(-4)+36 \\ =-64+64-36+36 \\ =0 \end{gathered}

Then (x + 4) is a factor. Let's do the synthetic division:

Then,

x^3+4x^2+9x+36=(x+4)(x^2+9)

Since

x² + 9 cannot be factor, we have completed our factoring:

\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \\ =(x-1)(x-2)(x+4)(x^2+9) \end{gathered}

Now, we have the following expression:

(x-1)(x-2)(x+4)(x^2+9)=0

Then, we have five posibilities:

(x - 1) = 0

or (x - 2) = 0

or (x + 4) = 0

or (x² + 9) = 0

Then, we have five solutions;

x - 1 = 0 → x₁ = 1

x - 2 = 0 → x₂ = 2

x + 4 = 0 → x₃ = -4

x² + 9 = 0 → x² = -9 → x = ±√-9 = ±√9√-1 = ±3i

→ x₄ = 3i

→ x₅ = -3i

<h2><em>Answer- the solutions of the polynomial are: x₁ = 1, x₂ = 2, x₃ = -4, x₄ = 3i and x₅ = -3i</em></h2>

7 0
11 months ago
...................... can yall answer this real quick pleaseee
KIM [24]

Answer:

5.7

Step-by-step explanation:

4 0
2 years ago
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Please I need help for number 6 for brainlest and answer ! And show your work on how you got it
solmaris [256]

Answer:

Step-by-step explanation:

inverse variation. if value of x increase, y value decrease.

For example,

If x=10;   y =1/10 = 0.1

If x =100 ;  y = 1/100 = 0.01

8 0
3 years ago
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Jynessa wants to order these fractions: 4/9, 2/3, 1/6, -2 1/2 What should she use as her common denominator?
IceJOKER [234]
4/9 , 2/3, 1/6, ....-2 1/2 = - 5/2

common denominator is 18
8 0
3 years ago
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