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Otrada [13]
3 years ago
7

Could i have some help please!!!!!!! 15 pts

Mathematics
1 answer:
Zigmanuir [339]3 years ago
5 0
Simple...

you have: 9u=10u-6

Isolate the variable-->>u

9u=10u-6

9u=10u-6
-10u  -10u

-1u=-6

Now simply divide-->>

u=6

Thus, your answer.


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x^2+y^2=9\Leftrightarrow (x-0)^2+(y-0)^2=3^2 hence the center is (0,0) and the radius is 3
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Which table can be created using the equation below? –2 + 4x = y A 2-column table with 3 rows. Column 1 is labeled x with entrie
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The correct answer is C. A 2-column table with 3 rows. Column 1 is labeled x with entries negative 5, 0, 3. Column 2 is labeled y with entries negative 18, negative 2, 10.

Explanation:

The purpose of an equation is to show the equivalence between two mathematical expressions. This implies in the equation "–2 + 4x = y" the value of y should always be the same that -2 + 4x. Additionally, if a table is created with different values of x and y the equivalence should always be true. This occurs only in the third option.

   x                 y

   5                -18    

   0                -2

   3                 10

First row:

-2 + 4 (5) = y  (5 is the value of x which is first multyply by 4)

-2 + 20 = -18 (value of y in the table)

Second row:

-2 + 4 (0) y

-2 + 0 = -2

Third row:

-2 + 4 (3) = y

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Answer:

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The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.

The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.

Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:

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This is zero when ...

  (r -4)^2 = 7

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Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...

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_____

<em>Comment on the graph</em>

I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)

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It would be circular since it was cut from a cylindrical log.

Hope This Helps You!
Good Luck Studying :)
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3 years ago
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