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1 year ago

Please find me the value of x

Mathematics

2 answers:

ipn [44]1 year ago

7 0

Answer:

x = 60°

Step-by-step explanation:

∠ ABO and ∠ CBO are a linear pair and sum to 180° , that is

∠ ABO + ∠ CBO = 180°

150° + ∠ CBO = 180° ( subtract 150° from both sides )

∠ CBO = 30°

∠ XYO and ∠ CBO are alternate angles and are congruent , so

∠ XYO = 30°

∠ XOY and ∠ COY are a linear pair and sum to 180° , then

∠ XOY + ∠ COY = 180°

∠ XOY + 90° = 180° ( subtract 90° from both sides )

∠ XOY = 90°

the sum of the 3 angles in Δ XOY = 180° , that is

x + 90° + 30° = 180°

x + 120° = 180° ( subtract 120° from both sides )

x = 60°

erastova [34]1 year ago

4 0

Answer:

x = 60°

Step-by-step explanation:

See picture below. When two parallel lines are cut by a transversal there are some special relationships.

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2 years ago

Read 2 more answers

I need help with this

andre [41]

Answer:

$f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}$

Step-by-step explanation:

Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.

Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².

Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).

Taken altogether, these factors give us the function ...

$\boxed{f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}}$