$\bf tan(\theta )=9\implies tan(\theta )=\cfrac{\stackrel{opposite}{9}}{\stackrel{adjacent}{1}}\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf c=\sqrt{1^2+9^2}\implies c=\sqrt{82} \\\\[-0.35em] ~\dotfill\\\\ cos(\theta )=\cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{82}}}\implies \stackrel{\textit{rationalizing the denominator}}{cos(\theta )=\cfrac{1}{\sqrt{82}}\cdot \cfrac{\sqrt{82}}{\sqrt{82}}}\implies cos(\theta )=\cfrac{\sqrt{82}}{82}$

3 0

3 years ago

Find the zeros using the quadratic formula x^2-5x+4=y.

Ber [7]

Answer:

(x-4) (x-1)

Step-by-step explanation:

X-4=0

x=4

X-1=0

X=1

8 0

3 years ago

What is the area of the figure? Please show work?

WARRIOR [948]

Answer:

Area of ΔABC = 144 ft^2

Step-by-step explanation:

ΔABC is 45 45 90 right triangle

Ratio of leg (AC, BC) : hypo (AB) = x : x√2

Given hypo AB = 24 so AC = BC = 12√2

Area of ABC = 1/2 bh

= 1/2 ( 12√2)( 12√2)

= 1/2(144 * 2)

= 144 ft^2

3 0

3 years ago

Is the following relation a function?

Leto [7]

no it is not one x cannot have 2 y's.

8 0

3 years ago

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