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When you’re looking for zeros of polynomials and you already have one of the results, you should try Horner’s method(/scheme/rule).
It works like this: write the coefficients of your polynomial in one row (ALL of them. For example, if one of the coefficients is 0, you will need to write that down, too.)
In your case, it would be 1 -3 1 1.
Now you would write the number you’re suspecting to be a zero in the next row, but to the left so it looks like a kind of table now.
Then you start calculating. You always add the numbers downwards, and to get to the next column you mutiply by your zero suspect. So it’d be
1 -3 1 1 -> top row: coefficients of your origional polynomial
1 / 1 -2 -1 -> middle row: zero suspect and calculation
1 -2 -1 0 -> bottom row: coefficients of your resulting poynomial
In case your zero suspect turn out to actually be a zero of your polynomial, the last number on the bottom right should be a 0. In our case, it is, so the bottom row now tell us which coefficients our resulting polynomial has.
We have now divided x³-3x²+x+1 by x-1 (the first zero) and got x²-2x-1.
You are right, the other two zeros aren’t whole numbers but you can use the regular methods for finding roots of quadratic equations. (you could, for example, check our tags …)
I hope that helped!
28 - (-(-15)) - |10| = ?
The double negatives cancel inside the bracket and the | | just means that the number inside becomes positive, since it’s already positive the absolute sign drops to make:
28 - (15) - 10 = ?
13 - 10 = 3
To find each intercept, you must let the other variable be equal to zero to find the point of the other axis.
For example, if you were to find the x intercept on this function you allow:
y = 0
3x + 4(0) = -12
3x + 0= -12
x = -4
and to find the y intercept you let x =0
3(0) + 4y = -12
0 + 4y =-12
y = -3
so you’re points of intercept are (-4,0) for the x intercept. and (0,-3) for the y intercept
Hi! Sorry for the delay in answering your question! Here’s a solution to the problem, I even did a little extra to show you how you can develop an idea of how the function would look when plotted.
If you need a higher quality picture try this one here
Re: the last post, simplifying (Tan[x+y] + Tan[x-y])/(1 - Tan[x+y]Tan[x-y]), the cheap way to simplify it is to remember the formula for a trig function of a sum/difference of angles:
Tan[x+y] = (Tan[x] + Tan[y])/(1 - Tan[x]Tan[y])
Which pretty much spits out the answer of Tan[2x] as the simplified version of the given expression. You can derive that formula for Tan[x+y] pretty easily from the more well-known trig addition formulas
Sin[x+y] = Sin[x]Cos[y] + Sin[y]Cos[x]
Cos[x+y] = Cos[x]Cos[y] - Sin[x]Sin[y]
And you can see proofs of these formulas here.
Are you sure it said 80 meters and not 80km? Because it’s far more likely for two reasons: a) i hardly believe exists something you can drive that goes 80m/h and b)if you work out a proportion, it gets a much more believable result
60minutes:80000meters = x minutes : 128748meters -> (60*128748):80000 gives you 96.561 minutes which are roughly 1.6 hours, which is believable if you think that you are doing 80.000 in 60 minutes.
Otherwise, if the question really asked for 80m/h, then your friend who said 1609.35 hours is right :)
(Also because, if you think about it, if you are doing 80meters in one hour, in 4 hours like your friend said you would have done 80*4meters=320meters which are roughly 0.2 miles)
Hiii idk if this has already been answered as im on mobile but my question is to simplify the following: (tan(x + y) + tan (x - y)) / (1 - tan (x + y) tan (x - y))
to answer that I worked with the definition of the tangens: tan(a) = sin(a)/cos(a). First I changed all tan’s into sin and cos. The numerator can then be summarised into one big fraction and when you add it you’ve got
[sin(x+y)cos(x-y) + sin(x-y)cos(x+y))] / cos(x+y)cos(x-y).
Now you can do the same thing to the demoninator; the product of the tans can be written as [sin(x+y)sin(x-y)]/cos(x+y)cos(x-y). To subtract that from 1 you have to write 1 as cos(x+y)cos(x-y) over cos(x+y)cos(x-y).
That doesn’t look much simplier but now you’ve got double fractions you can cancel the denominator cos(x+y)cos(x-y) from both parts. The result you get is
[sin(x+y)cos(x-y)+sin(x-y)cos(x+y)] / [cos(x+y)cos(x-y) - sin(x+y)sin(x-y)]
I then divided by sin(x+y) so you get
[cos(x-y)+sin(x-y)cot(x+y)] / [cot(x+y)cos(x-y) - sin(x-y)]
since the co-tangens is defined as 1/tan and therefore cos/sin. Now I divided by cos(x-y) to get
[1+tan(x-y)cot(x+y)] / [cot(x+y) - tan(x-y)]
It’s basically just an alternative from the form you had at the beginning and thus I’m not really sure if this was the result you were supposed to get. There are a lot of alternatives from the big fraction: you could, for example, divide by sin(x+y) but then by sin(x-y) to get
[cot(x-y)+cot(x+y)] / [cot(x+y)cot(x-y) - 1]
and so on. So I’m not sure if that was helpful - if anyone else has a better idea, please reply/inbox/submit/etc.
A continuation of the bearings question:
Sorry about not specifically mentioning what methods the book had in mind. The idea that you obviously don’t have the book in front of you kind of flew over my head. Given I don’t have the most confident grip on bearings yet, I’m just going to copy what the text says here.
Method 1: When a single angle is given, such as 164, it is understood that the bearing is measured in a clockwise direction from due north.
Method 2: The second method starts with a north-south line and uses an acute angle to show direction, either east or west, from this line.
I think where I get confused about why there isn’t a south measurement is that the second method requires a ‘north-south line’. So why wouldn’t a south measurement be involved in the aforementioned problems? Thanks again for your continued help and patience!
I got it now, thanks for submitting again. And, curiously, I already explained why you don’t have a south measurement from those two points: they lie in the north. That’s it.
You wrote N0°W (or N0°E) for (0,4) and N45°W for (-5,5). So, for example, if you now had (0,-4) you’d get S0°W (or S0°E) and for (-5,-5) you’d get S45°W. But you’re in the range of the positive y-axis, so you’re only in the north. I hope that helped!
Hello, I have a couple of specific questions and a general question. Currently, I’m in the process of learning ‘bearings’ in trigonometry and in general I don’t think I understand the concept despite several rereads of my textbook. I must get something, as I get my problems half right [as shown above in blue and purple]. However, I don’t understand how, in example either problem above, there is no South measurement. Due to not understanding this I’ll usually add a south measurement and get the problem wrong. The exact prompt that goes the with the blue and purple problems is: ‘An observer for a radar station is located at the origin of a coordinate system.For each of the points in these exercises, find the bearing of an airplane located at that point. Express bearing using both methods.’
Aside from that, there’s a question shortly after that that I am completely lost on. I’ve never been good at word problems, so I don’t even know where to begin.
The ray y=x, x is greater than or equal to 0, contains the origin and all points in the coordinate system whose bearing is 45 degrees. Determine the equation of a ray consisting of the origin and all points whose bearing is 240 degrees.
The given solution to this last problem is in red.
Thank you so much for your help!
For the first paragraph: as far as I understood, there is no South measurement required here. The reason why you’ve only got North measurements is that the points (-5,5) and (0,4) are on the upper side of the coordinate system. Or could you explain what are the both methods the prompt refers to?
The second questions requires some meddling with angles. I’ve invested some minutes into Paint so let’s look at the graphic:
The blue line represents the angle of 240° we know we have. The red line is the final equation (just like the red result in your picture) with the arrow showing which direction the airplane would go.
Now, if we want an equation in an x-y-coordinate system, we have to change the way we measure angles. When talking about bearings, you take the North/positive y-axis and go clockwise. But when you’re working with an x-y-coordinate system, the angle is measured anti-clockwise starting from the positive x-axis (this direction is also called “mathematically positive”). This very angle is orange in the graphic.
The equation we’re looking for is a linear one, thus it has the form of y=m x + n. Since it passes through the origin, n=0. Furthermore, m=tan(alpha), wherein alpha is the orange angle.
Back to the graphic. We can see that the blue angle is more than 180°, so when we subtract 180° we get the turquoise angle, which is, in this case, 60°. It’s also clear that the angle that adds up with the turquoise one to 90° is the opposite angle of our orange one. And since opposite angles always have the same value, our orange angle is 90°-60° = 30°.
When we put those 30° into a tangens we get the exact gradient from the solution you got in your graphic. The part of this equation they want is the one where the bearing is 240° - so I dashed the part that is not asked for. You can express this by writing “for x [smaller and equal to / <=] 0” like in the solution.
I hope that helped - if you need any more help, don’t hesitate to mail us!
Map of Mathematistan
Lots of hidden puns. Notice for instance how the houses in Arithmetics are arranged according to the Fibonacci sequence, Gödel’s incompleteness theorems as a whirlpool in the Ocean of Logic, the normal distribution statue in Statistigrad, the peakless Mt. Infinity, and plainly fields.