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Kaylis [27]
2 years ago
14

Can someone please help me with this geometry question I will mark brainliest

Mathematics
2 answers:
Papessa [141]2 years ago
5 0

Answer:

hypotenuse = 8\sqrt{3} , leg = 12

Step-by-step explanation:

Using the cosine and tangent ratios in the right triangle and the exact values

cos60° = \frac{1}{2} , tan60° = \sqrt{3} , then

cos60° = \frac{adjacent}{hypotenuse} = \frac{4\sqrt{3} }{hypotenuse} = \frac{1}{2} ( cross- multiply )

hypotenuse = 4\sqrt{2} × 2 = 8\sqrt{2}

---------------------------------------------

tan60° = \frac{opposite}{adjacent} = \frac{opposite}{4\sqrt{3} } = \sqrt{3} ( multiply both sides by 4\sqrt{3} )

opposite = 4\sqrt{3} × \sqrt{3} = 12

likoan [24]2 years ago
4 0

Answer:

Lengths are 8√3 and 12

Step-by-step explanation:

» From trigonometric ratios, using our angle as 60°

{ \tt \cos( \theta) =  \frac{adjacent}{hypotenuse}  } \\  \\  { \tt{ \cos( 60 \degree)  =  \frac{4 \sqrt{3} }{hypotenuse} }} \\  \\ { \tt{hypotenuse =  \frac{4 \sqrt{3} }{ \cos(60 \degree) }  }} \\  \\ { \tt{hypotenuse =  \frac{4 \sqrt{3} }{0.5} }} \\  \\ { \boxed{ \tt{hypotenuse = 8 \sqrt{3} }}}

» Using 30° as our angle:

{ \tt{ \tan( \theta)  =  \frac{opposite}{adjacent} }} \\  \\ { \tt{ \tan(30 \degree)  =  \frac{4 \sqrt{3} }{adjacent} }} \\  \\ { \tt{adjacent =  \frac{4 \sqrt{3} }{ \frac{1}{ \sqrt{3} } } }} \\  \\ { \tt{adjacent =  \frac{(4 \sqrt{3} ) \times ( \sqrt{3}) }{1} }} \\  \\ { \tt{adjacent = 12}}

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Type the correct answer in each box. Use numerals instead of words.
Natali5045456 [20]

Given:

Consider the given function is f(x)=-(x+1)(x-3)(x+2).

To find:

The remaining zero and y-coordinate of y-intercept.

Solution:

We have,

f(x)=-(x+1)(x-3)(x+2)

For zeros, f(x)=0.

-(x+1)(x-3)(x+2)=0

(x+1)(x-3)(x+2)=0

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x=-1,x=3,x=-2

So, three zeros of given function are -1, 3 and -2.

Putting x=0 in the given function, we get

f(0)=-(0+1)(0-3)(0+2)

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f(0)=6

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3 years ago
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