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3 years ago

The set of possible values of m is (5, 7, 9).

Mathematics

1 answer:

k0ka [10]3 years ago

4 0

Plug in each possible value into the equation.

when m = 5

2k = m + 3

2k = (5) + 3

Simplify

2k = 8

isolate the k, divide 2 from both sides

2k/2 = 8/2

k = 8/2

k = 4

When m = 5, k = 4

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2k = m + 3

when m = 7

2k = (7) + 3

2k = 10

2k/2 = 10/2

k = 10/2

k = 5

when m = 7, k = 5

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2k = m + 3

When m = 9

2k = (9) + 3

2k = 12

2k/2 = 12/2

k = 12/2

k = 6

when m = 9, k = 6

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hope this helps

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Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4

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Answer:

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

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solution:

$\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}$

use <em>(a - b)(a + b) = a² - b²</em>

$\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}$

multiply both sides by (x - 3) ≠ 0

$\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}$

cancel (x - 3)

$\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3$

subtract $\frac{4(x-3)}{x}$ from both sides

$\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}$

cross multiply

$(4)(x)=(x+3)(-x+12)$

use FOIL

$4x=(x)(-x)+(x)(12)+(3)(-x)+(3)(12)\\\\4x=-x^2+12x-3x+36$

subtract 4x from both sides

$0=-x^2+12x-3x+36-4x$

combine like terms

$0=-x^2+(12x-3x-4x)+36\\\\0=-x^2+5x+36$

change the signs

$x^2-5x-36=0\\\\x^2-9x+4x-36=0\\\\x(x-9)+4(x-9)=0\\\\(x-9)(x+4)=0$

The product is 0 if one of the factors is 0. Therefore:

$x-9=0\ \vee\ x+4=0$

$x-9=0$ <em>add 9 to both sides</em>

$x=9\in D$

$x+4=0$ <em>subtract 4 from both sides</em>

$x=-4\in D$

7 0

3 years ago

Find the value of c so that (x-5) is a factor of the polynomial p(x)

liubo4ka [24]

I think the question is

Find the value of c so that (x-5) is a factor of the polynomial

$p(x) = x^3 + 2x^2 + cx + 10$

The other factor is going to be some quadratic. We can say a few things about its coefficients but let's start by saying in general it's

$q(x)= ax^2 + bx + k$

$p(x) = (x-5)q(x)$

$x^3 + 2x^2 + cx + 10 = (x-5)(ax^2 + bx+k) = ax^3 + (b-5a)x^2 + (k-5b)x - 5k$

Equating respective coefficients,

$a=1$

$b-5a = 2$

$k - 5b = c$

$-5k = 10$

so we get

$b = 2 + 5 = 7$

$k = 10/-5 = -2$

$c = k - 5b = 2 - 5(7)= -37$

Answer: -37

Check:

$(x^2 + 7x - 2)(x - 5) = x^3 + 2 x^2 - 37 x + 10\quad\checkmark$

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