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3 years ago

Dx+hy=j,forx x= thanx for any and all help

Mathematics

2 answers:

Crazy boy [7]3 years ago

6 0

$dx+hy=j \\ \\ dx = j-hy \ \ / : \ d \\ \\d\neq 0\\ \\x=\frac{j-hy }{d}$

Marianna [84]3 years ago

6 0

Dx + hy = j

Dx = j - hy

x= (j - hy) / D

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I can’t figure out how to use the zeros in the polynomial. Please explain

Temka [501]

Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

c) $P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

Step-by-step explanation:

For the question a * you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

For question b, do the same procedure.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

---------------------------------------

$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

--------------------------------------

$x = 6\\\\(x-6) = 0$

--------------------------------------

$P (x) = (2x + 5) (5x - 4) (x-6)$

Finally for the question c we have

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

$x = -3\\\\(x-3) = 0$

--------------------------------------

$x = 1\\\\(x-1) = 0$

--------------------------------------

$x = 4\\\\(x-4) = 0$

----------------------------------------

$x = -1\\\\(x + 1) = 0$

-----------------------------------------

$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

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3 years ago

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