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For an angle Θ with the point (−20, −21) on its terminating side, what is the value of cosine?

Mathematics
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@kelliegirl33 can you help me
If it helps you graph the point. Can you do that for me first?
@myininaya just plot -20,-21 ?

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Other answers:

Yep plot the point as a first step.
You can use the draw tool.
ok
|dw:1403474666905:dw|
|dw:1403474755600:dw| Now I will simply make a right triangle using the center of the graph for my central angle.
what is the ratio definition of cosine of an angle?
legnth of adjacent / legnth of hyp
right you already have the adjacent side given in the picture the only thing you need now is the hyp of that triangle, right?
yes
Find the hyp of that triangle using Pythagorean Theorem.
the adjacent side is -20
You are correct to say that.
Can you find the length of the hyp?
yes hold on wouldnt it be -20^2 + 21^2 = c^2
20^2+21^2=c^2 and then take the square root of both sides to give you c.
29
so that is the length of the hyp so what is cos(theta)=?
29 ?
you said earlier cos(theta) is adj/hyp you never said cos(theta) is hyp
20/29
don't forget the negative sign
thank you so much i have a few more cn yu help me
if the angle was on the side of the graph like on the right hand side of the y-axis cos would have been indeed a positive number but since the angle was to the left of the y-axis cos will be negative
|dw:1403475467323:dw|
How does changing the function from f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function?
we could even talk about sine sin is positive when above the x-axis and sine is negative when below the x-axis |dw:1403475522177:dw|
Well you are talking about shifting the graph up when you have that +5
so the function shifts up 5 units so the range changes from -1 to 1
in f(x) to 4 to 6 in g(x)
For example, pretend we have y=sin(x) the range of sin(x) is from -1 to 1 (inclusive). but when we look at y=sin(x)+1 that means you are moving every point of y=sin(x) up one unit so the range of sin(x)+1 is from -1+1..1+1( inclusive) doing the addition gives 0..2 (inclusive)
am i right
well first what is the range of y=3sin(x)?
f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function
yes the range of the graph of f would be affected if you need f+5
did f+5 (not need)
but the range of y=3sin(x) is not -1 to 1.
Which of the following functions has the greatest y-intercept?|dw:1403475803380:dw|
its 2 to 8
Let y=sin(x). Let f(x)=3sin(x). Therefore you could say f(x)=3y where y=sin(x). If you have 3 times every y per x wouldn't the range be stretched more?
yes
like the range for y=sin(x) is -1 to 1 (when we talk about range we are talking about the y values) the lowest y value for y=3sin(x) would be what then?
so f(x) has greater y-intercerpt
the lowest y value for y=sin(x) is -1 so the lowest y value for f(x)=3sin(x) is ?
-3
great so it shouldn't be hard for you to see the highest y-value for f(x)=3sin(x) would be 3(1)=3 so the range for y=3sin(x) is -3 to 3 (inclusive)
now what is the range for y=3sin(x)+5
2 to 8
omg you are great so now you can answer that one question about how the range is affected and actually state the ranges of both functions
now for Which of the following functions has the greatest y-intercept? wouldnt it be f(x) since the y is 9
Also when answering the question it might be better to choose to answer in interval notation or inequality notation. 2 to 8 isn't really something people say to state an interval (not that I have ever seen anyways)
Okay , so it would be stated has 2 to 8 in g(x)
well like i said you might want to choose to answer in interval notation (or inequality) y is an element of the interval [2,8] <--interval (i put brackets to mean to included endpoints inequality notation: \[2 \le y \le 8\] and to your next question.... |dw:1403476409004:dw| do you what the y-intercept is for a curve? It is the part of the curve that goes through the y-axis. When I say part, I'm talking about the specific point in which the curve does that. What do you know about all the points on the y-axis? What do they all have in common?
I will give you a hint. All the x values of any point on the y-axis are _____?___
Im not sure
4-9 = -5 which is the same for g(x)
the same y-intercept
Ok since you can't answer that, I will ask you a series of questions now? Where does (0,10) lie? Where does (0,5) lie? Where does (0,2) lie? Where does (0,0) lie? Where does (0,-1) lie? Where does (0,-6) lie? where does (0,-1000) lie?
on the x axis
That is incorrect. They all lie on the y-axis. The answer to my question: All the x-values of any point on the y-axis is ____?__ All the x-values of any point on the y-axis is zero. So to find the y-intercept for a curve you just need to find y when x is zero. You just need to replace x with zero and solve for y.
so 0-9 ? 0-4 ?
For the one function, it should be clear (0,0) is the y-intercept (that is the origin of the graph after all; is is all an x-intercept) The other function you did right even though it doesn't seem you know why you were doing it since you were not able to answer the question.
Yeah im confused now
|dw:1403477105487:dw| This is the information given. You are asking to find the y-intercept for both graphs right? Can you tell me which point from f have the x is 0?
functions has the greatest y-intercept
1,1 ?
I'm asking you to tell which of the points from f have the x value is 0?
|dw:1403477214481:dw|
its 0,0 ?
|dw:1403477221682:dw| Do you see any of these x's being 0?
yes
yes the point (0,0) is a y-intercept because the x is 0
now we need to look at g(x)=5cos(3x)-5
what point on g gives us the x is 0?
-5
how did you get that?
Did you replace the x with 0 to see what y is? Because y is not -5 when x is 0.
I only care about when x is 0 since you are asking about the y-intercept.
Replace the x with 0 like so y-intercept of g is g(0)=5cos(3*0)-5
evaluate 5cos(3*0)-5
5cos(0)-5 5cos
you are right to say 5cos(3*0)-5 can be wrriten as 5cos(0)-5 but where do we go from there?
1.41
:( 5cos(0)-5 what is cos(0)?
1
5(1)-5 can you simplify from here now?
0
same y
right so the y-intercept for g is (0,0) as well they have the same y-intercept
1f sin Θ = 15 over 17, use the Pythagorean Identity to find cos Θ.
draw a right triangle using this information sin(theta)=15/17
\[\sin^2 \theta + \cos ^2\theta = 1\] will be what we use ?
you can use that
|dw:1403477725033:dw|
|dw:1403477767271:dw|
if you want to use that specific identity you can replace sin(theta) with 15/17 and solve for cos(theta)
sin(15) = 0.650
sin*17) = -0.96
can you replace sin(theta) with 15/17 in the following equation \[\sin^2(\theta)+\cos^2(\theta)=1\]
no
why not?
well wouldnt you put them in front of sin and cos
just so you know \[\sin^2(\theta)\] and \[(\sin(\theta))^2\] have exactly the same meaning
you are given sin(theta) is 15/17 so you replace the whole sin(theta) thing with 15/17
\[\sin^2(\theta)+\cos^2(\theta)=1 \text{ means exactly that } (\sin(\theta))^2+(\cos(\theta))^2=1\] So we have \[(\sin(\theta))^2+(\cos(\theta))^2=1 \\ (\frac{15}{17})^2+(\cos(\theta))^2 =1 \text{ I did this because I was given that } \sin(\theta)=15/17\]
now solve for cos(theta)
if that confuses you solve the following equation for u \[(\frac{15}{17})^2+u^2=1 \]
whatever value(s) you get for u is your answer since u is cos(theta)
and cos(theta) is the objective
225/285
|dw:1403478332917:dw|
so what did you do to get that?
found cos of theta
you solve the equation (15/17)^2+u^2=1 for u (which is cos(theta) )?
yes
you might want to do that one more time
the denominator is right but your numerator is off
8 / 17
ok so cos could either be 8/17 or -8/17 when sin is 15/17
|dw:1403478613956:dw| my only option
Agreed
In Shadow Bay in July, high tide is at 1:30 pm. The water level is 4 feet at high tide and 0 feet at low tide. Assuming the next high tide is exactly 12 hours later and the height of the water can be modeled by a cosine curve, find an equation for Shadow Bay's water level in July as a function of time (t).
okay so i got f(t) = 4cos pie/2t +1

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