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nexus9112 [7]
3 years ago
8

I'm being timed pls help will pick brainliest

Mathematics
1 answer:
stiv31 [10]3 years ago
7 0

Answer:

Cos 0° = 6.42 = V1

6th quadrant.

So, cos x = 0 implies x = (2n + 1)π/2 , where n takes the value of any integer. For a triangle, ABC having the sides a, b, and c opposite the angles A, B, and C, the cosine law is defined. In the same way, we can derive other values of cos degrees like 30°, 45°, 60°, 90°, 180°, 270°and 360

Step-by-step explanation:

if sin a= 3/4 then a = 50

if sin = 4/5 then a = 60

if sin = 2/3 then a = 40

But we can perfect this

if sin = 4/5 then a = 57   as 4/5 = 0.83

We want 0.8 = 4/5

if sin = 4/5 then a = 54 as 4/5 = 0.809

if sin = 4/5 then a = 53.5 as 4/5 = 0.80385

Now for cos

It is much easier than it initially appears. Remember the definition of SINE:

SINΘ = opp             In your case, that means the opposite is 4/5 = 0.80385 (yes, ignore the sign for now) and the hypotenuse is 11.48910018

           hyp             Please draw that triangle right now, because it will help you a lot at the end.

 53.50 degree           Remember to place the angle in the appropriate spot.

Now, use the Pythagorean Theorem to find the missing side (easy, right? It's 9.534) and place it in the adjacent position.

You can easily find all of the trig functions now!

Simply remember that:

COSΘ = Adj                                  with SEC the reciprocal of this one

            Hyp

TANΘ = Opp                                  COT the reciprocal of TAN and, if anyone asks, CSC coming from SIN.

            Adj

I told you to ignore the signs, but now we can't anymore. Remember the four quadrants and the memory trick:

A                        -- ALL are positive

Smart                 -- SIN and CSC are positive

Trig                    -- TAN & COT are positive

Class                  -- COS and SEC are positive.

Since your SIN was negative, it must be in III or 6, and COS is positive in I and 6 So we're in quadrant 6 then!

Only your COS and SEC will be positive, the rest negative

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Identify an equation in point-slope form for the line perpendicular to y=-1/3x-6 that passes through (–1, 5).
Fittoniya [83]
<span>y=-1/3x-6 slope = -1/3
</span><span>perpendicular  lines, slope is opposite and reciprocal
</span><span>
so slope = 3
</span><span>passes through (–1, 5).
</span>
Point-slope<span> is </span><span>point-slope form </span><span>y-y₁=m(x-x₁)
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4 0
3 years ago
Hello need help with this one to and thanks and have a great day or weekend and be safe
Mandarinka [93]

Answer:

a) It's a quadratic

b) f(x) = A(x-3)(x-7)

c) she is right

Step-by-step explanation:

(0-12)/(3-1) = -6

(-4-0)/(5-3) = -2

Since the slope is not constant, it can't be a linear.

Therefore a quadratic

Since the zeroes are 3 and 7,

Factors are (x - 3) and (x - 7)

f(x) = A(x - 3)(x - 7)

Use any one point besides the x-intercepts to find A

12 = A(1-3)(1-7)

12 = 12A

A = 1

f(x) = (x - 7)(x - 3)

f(x) = x² - 7x - 3x + 21

f(x) = x² - 10x + 21

She is right

7 0
3 years ago
Read 2 more answers
I need help with this any help will be appreciated
Mars2501 [29]
Are you trying to solve for x and y?

y=2x+1; x+2y=17
sub in one variable in the other equation.
x+2(2x+1)=17
x+4x+2=17
5x=15
x=3

now that you have x value, plug into the original equation to find y and check with the other. 
y=2(3)+1 = 7
x+2y=17 = 3 + 2(7) = 17 ; 17=17 
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