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Brrunno [24]
3 years ago
10

Ella S, Lily, Aziza, and Kevin earned a total of $100 shoveling snow. Each of them earned the same amount. How much did each of

them earn? I will give 13 points
Mathematics
1 answer:
Lesechka [4]3 years ago
3 0

Answer:

$50 :)

Step-by-step explanation:

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ASAP Use elimination -2x+5y=-20 <br> 2x-2y=20
Inga [223]

Answer:

X=10, Y=0

Step-by-step explanation:

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A dance academy charges $10 per class and a one-time registration fee for $50 , a student pays a total of $90 to the academy , f
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Answer:4 classes

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3 years ago
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
1 year ago
SHOW ALL WORK BUT KEE IT SIMPLE AND EASY TO UNDERSTAND:)
7nadin3 [17]
To do this problem, you need to use a process called completing the square. Let me explain:  

To complete the square on the function f(x) = x² + 8x +13, first group the first two terms in (  )  and leave some space at the end as follows:
f(x) = (x² + 8x          ) + 13  Now our next step is to fill in the space and adjust our expression on the right hand side of the function.  To do this, we take half of the middle number 8 and then square it:  so 4² = 16 and we fill in our space inside the ( ) with this value 16;   
f(x) = (x² + 8x + 16) + 13  now what we have done is to increase the overall value of our expression on the right by 16, but we want the overall value to remain the same.  To fix this we simply need to subtract 16 at the end like this:  f(x) = (x² + 8x + 16) + 13 -16   we can simplify and get the following.
f(x) = (x² + 8x + 16) - 3   At this point we're almost done.. All we need to do now is to rewrite the what is in the parentheses in a slightly different form.  Here is what it will look like:  f(x) = (x + 4)² - 3  notice all I did was take the sum of the square root of x² and the square root of 16 originally in the (  ) to get then new expression inside the ( ) and then square that  ( )²

Now this is a nice form to have because you can get the vertex straight from this form.. IN FACT this is called vertex form or (h,k) form for short.  In general the form is f(x) = a(x - h)² + k  don't worry about the 'a'  for now.. you might see that in our case it is just 1 and will not effect our equation.  You only have to consider this if the original leading coefficient of the quadratic is not 1 to begin with...

So you can see that our vertex is (-4,-3)  
Hope this is helpful, but if you have questions let me know.
5 0
3 years ago
Solve the proportion.4/12=x/18
juin [17]
Solve for x by cross multiplying and the answer is x=6


hope this helps and plz rate below :)
5 0
3 years ago
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