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egoroff_w [7]
3 years ago
12

The perimeter of a college athletic field is 104 meters and the length is 16m more than the width find the length and width

Mathematics
1 answer:
Karolina [17]3 years ago
8 0
Attached a solution and showed work.

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15/25 in lowest terms
klio [65]
3/5 is the lowest term. 3/5 = 15/25. Does this help? 
8 0
3 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
1 year ago
If g(x) = 2x − 1, what is g(−2.3)?
Kaylis [27]

Answer:

Gx 2x -1 (2,-5)

Step-by-step explanation:

I think that was the thing on apex sorry if wrong

3 0
3 years ago
What is the interquartile range of this set of data? 15, 19, 20, 25, 31, 38, 41
Irina18 [472]
First, you have to find the median, which is the number in the middle: 25

Then cut the data in half by the median, so your new data sets are
15,29,20
    and 
31,38,41

Now you find the median of both of those sets, which is 29 and 38.

The interquartile range is the difference between the numbers, so 38-29 = 9.
8 0
4 years ago
Solve<br> x2 - 49 = 0<br><br> A) ±49 <br> B) ±12 <br> C) ±7 <br> D) ±6
Alex Ar [27]
C, x = +-7. Difference of squares
5 0
3 years ago
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