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lawyer [7]
3 years ago
13

PLEASE HELP ME WITH BOTH QUESTIONS THANKS!!

Mathematics
1 answer:
vitfil [10]3 years ago
8 0

First one:

cos(A)=AC/AB=3/4.24

cos(B)=BC/AB=3/4.24

Cos(A)/cos(B)=AC/AB / (BC/AB) = AC/AB * AB/BC = AC/BC=3/3=1


Second one:

To solve this problem, we have to ASSUME AFE is a straight line, i.e. angle EFB is 90 degrees. (this is not explicitly given).

If that's the case, AE is a transversal of parallel lines AB and DE.

And Angle A is congruent to angle E (alternate interior angles).

Therefore sin(A)=sin(E)=0.5

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5/x+3 take away 1/x-2 equals to 1. solve two possible answers for x
satela [25.4K]

Answer:

<h3><u>Solution</u><u>:</u><u>-</u></h3>
  • First write the given equation

\qquad{:}\longrightarrow\sf \dfrac {5}{x+3}-\dfrac{1}{x-2}=1

\qquad{:}\longrightarrow\sf \dfrac {5 (x-2)-1 (x+3)}{(x+3)(x-2)}=1

\qquad{:}\longrightarrow\sf \dfrac {5x-10-x }{x^2-6x-6} =1

  • use cross multiplication method

\qquad{:}\longrightarrow\sf 5x-x-10=x^2-6x-6

\qquad{:}\longrightarrow\sf x^2-6x-6=4x-10

\qquad{:}\longrightarrow\sf x^2-6x-4x-6+10=0

\qquad{:}\longrightarrow\sf x^2-10x+4=0

  • use quadratic formula

\qquad{:}\longrightarrow\sf x=\dfrac {-b\underline{+}\sqrt {b^2-4ac}}{2a}

\qquad{:}\longrightarrow\sf x=\dfrac {10\underline{+}\sqrt {(-10)^2-4×1×4}}{2×1}

\qquad{:}\longrightarrow\sf x=\dfrac {10\underline{+}\sqrt{100-16}}{2}

\qquad{:}\longrightarrow\sf x=\dfrac {10\underline{+}\sqrt {84}}{2}

\qquad{:}\longrightarrow\sf x=\dfrac {10+\sqrt {84}}{2}\quad or\quad x=\dfrac {10-\sqrt{84}}{2}

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