Yeah........I still have to wrap my brain around it.....LOL!!!!
What about finding the range for f(x)=-2x+5 with domain {1,2,3,4} ????
Thanks Chris, I had to think about it a little bit too :)
Hi asinus,
Your answer does not make sense.
x is not an angle it is a ratio. so you cannot have sin(x)
Certainly if the angles are both 45 degrees this will work
but one angle might be 30 degrees and the other 60degrees
or as I have said, one is x degrees and the other is 90-x degrees.
Think of this triangle and it might help you understand. :)
Note that all we really need to do is to find 2pi * 200 ≈ 1256.64 km
The radius of the first satellite = R1
The radius of the second satellite = R1 + 200
So......orbit of second satellite - orbit of first satellite =
2pi [R1 + 200] - 2pi [ R1] = 2pi [ R1 + 200 - R1] = 2pi *200
Points (a,5) and (9,b) are on the circle x2+y2=125. what are the values of a and b?
a^2 + 5^2 = 125 9^2 + b^2 = 125
a^2 + 25 = 125 81 + b^2 = 125
a^2 = 100 b^2 = 44
a = ± 10 b = ± √44
So...the points are (10, 5), (-10,5), (9, √44), ( 9, -√44)
For the equation of a circle the 2250000 = r^2 so r1 = 1500 r2=1700 km
Circumference for r1 = 2 x pi x 1500 km =9424.78 km
r2 = 2 x pi x 1700 = 10681.42
Difference r2 orbit is 1256.64 km longer in one orbit
No need to.....just be sure you're in degree mode [ select at bottom left ]
Connor might be an idiot.....but.....at least he doesn't misuse "your" for "you're "
ok, you got a gradient of about -16. It is definitely negative anyway because it slopes to the top left corner, so which anwer/s do younow know are definitely wrong?
Which answers are now still possibly correct?
What might the y intercept be approximately? Which answers does this rue out?
What answers are left now?
HINT:
remember that the equation of a line is y=mx+b
where m is the gradient and b is the y intercept
Thanks, Melody.....I couldn't think how to do this one....but ....I see what you did there....!!!!
cubrt (54) = cubrt(27*2) = 3 cubrt 2
cubrt(16) = cubrt (8*2) = 2 cubrt 2
add them together 5 cubrt(2)
∛54 + ∛16 =
∛[27 *2 ] + ∛[8 * 2] =
∛27 *∛2 + ∛8 *∛2 =
3∛2 + 2∛2 =
5∛2
What is the gravitational force of attraction of two asteroids that have a mass of 1,000,000kg and 75,000,000kg if they are 1000m apart?
\(F=G\frac{m_1\times m_2}{r2}\)
\(G=6.672\times10^{-11}\times\frac{m^3}{kg\times s^2}\)
\(\large F=6.672\times10^{-11}\times\frac{m^3}{kg\times s^2}\times\frac{10^6kg\times 75\times10^6kg}{10^6m^2}\)
\(F=0.005004\frac{mkg}{s^2} [N]\)
!
F = Gm1m2/r2 G = gravitational constant = 6.67 × 10-11 N-m2/kg2
So let's plug in the numbers
6.67x 10^-11 x 1 x10^6 x 75 x 10^6 / (1 x10^3)^2
= 500.25 x 10^(-11+6+6-6) = 500.25 x 10^-5 N
Well when i was waiting for your response, i did and tried the probelm myself, first, drawing the line of best fit, choose two good y-intercepts, did the m=y2--y1/x2-x1 and got -16. After that the probelm started to become weird. I chose a piar of coordinates and use it in the equation y=mx+b, pluging in the numbers and solving it. The answer i got was y=-16x+1324, which was sort of close to one of the answer choses.
The graph is here https://www.meta-chart.com/share/untitled-10071
Question is: Which equation best represnt the line of best fit?
A. y=-16x+400
B. y=-15x+300
C. y=15x+-300
D. y=16x+400
sin(arcsin x + arccos x)
\(sin(arcsin x + arccos x)\\ let\;\;asin(x)=\theta\\ then\;\;acos(x)=90-\theta\\ so\\ sin(arcsin x + arccos x)=sin(\theta+90-\theta)=sin(90)=1\)
I don't intend to give you just the answer. I am not here to do your homework but I am happy to help you learn.
*Is the gradient of this line positive or negative. Which choices are ruled out?
*What is the approximate y intercept? Which choices are ruled out?
Which choices are left?
Now by visual inspection, which is the best choice :)
If you want to interact with me I will help you work through it :)
THE CREEK INVENTED IT TO P**S OFF THE REST OF THE WORLD
Hi Sedadejo :)
I cannot speak Spanish but I do know how to use 'Google Translate' :) What is your mathematics question?
No puedo hablar español pero sí sé cómo usar 'Google Translate' :) ¿Cuál es su pregunta de matemáticas?
Area = pi r^2
r^2 = area/pi
r = sqrt(area/pi)
Circ = 2 pi r = 2 pi \(\sqrt{area/pi}\) = 2 sqrt(area x pi^2 /pi ) = 2 \(\sqrt{area*pi}\)
!
