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lana [24]
3 years ago
6

PLEASE HELP ME WITH THIS QUESTION FOR MATH!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
mezya [45]3 years ago
7 0

Answer:

∠MON = 130º

Step-by-step explanation:

∠LON = ∠LOM + ∠MON

∠LON = 180º

180 = (4x + 30) + (8x+ 90)

combine like terms

180 = 12 x + 120

60 = 12x

x = 5

plug x into equations.

∠MON = 8(5) + 90

∠MON = 40 + 90

∠MON = 130º

check work:

∠LOM = 4(5) + 30

∠LOM = 20 + 30

∠LOM = 50

∠LON = ∠LOM + ∠MON

180 = 50 + 130

kozerog [31]3 years ago
5 0

Answer:

MON = 130°

Step-by-step explanation:

LOM + MON = 180°

In other words

4z + 30° + 8z + 90 = 180°

12z + 120 = 180

60 = 12z

z = 5

5 x 8 = 40 + 90 = 130°

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3 years ago
Consider the following division of polynomials.
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x^2(x^2+2x+8)=x^4+2x^3+8x^2

Subtracting this from the numerator gives a remainder of

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-x^3=-x\cdot x^2. Multiplying the denominator by -x gives

-x(x^2+2x+8)=-x^3-2x^2-8x

and subtracting this from the previous remainder gives a new remainder of

(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8

This last remainder is exactly the same as the denominator, so x^2+2x+8 divides through it exactly and leaves us with 1.

What we showed here is that

\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}

=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}

=x^2-x+1

and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)

=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)

=x^4+x^3+7x^2-6x+8

which matches the original numerator.

3 0
3 years ago
Read 2 more answers
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