ah, 0.9pi is equal to 162 degrees right? i graphed it using my calculator. don't know why, but i thought of 0.9pi as 2.872433.
so would my equation look like this...
\int 4cos(x) -3sin(x) (a=0,b=.295\pi) + \int 3sin(x) - 4cos(x) (a=.295\pi,b=9\pi) is that correct?
Find the area of the region enclosed between y=3sin(x) and y=4cos(x) from x=0 to x=0.9\pi .
does that mean the integral would like this: \int 4cos(x) - 3sin(x) (a=0, b = 0.9*pi) is that correct?
k checked that part and got the same signs, so what is wrong with my answer? did i set up the equation correctly? (is there a Ti-83 plus program i can use to check my answer? or is there any math program for the computer that would do this for me?
Sketch the region enclosed by x+y^2=12 and x+y=0. Decide whether to integrate with respect to x or y. Then find the area of the region.
first thing i did was solve for x for each equation then set them to each other and got:
12-y^2=-y
y^2-y-12=0
found the points of intersection of y...
please help me check my answers, cause if i get one answer wrong, it's a big point deduction. i don't really like doing my homework online.
here it goes...
1.) An object cannot remain at rest unless which of the following holds?
A.)The net force acting on it is zero.
B.) The net...
sorry to go back to a question you already answered, but how did you find the time it will take?
i found that
V(0)x = 3.5m/s
V(0)y = 0 m/s
would you use this to find the time?
x(0) would be 4.5 since it's the initial height right?
x(t) = x(0) + v(0)t + 1/2(a)(t)^2?
0 = 4.5 + 3.5t +...
\int \frac{x^2}{\sqrt{(7x^2 + 8)}}
ok this one is a bit hard...i been thinking about this one for awhile.
can i do \int \frac{x^2}{\sqrt{(7x^2 + 7 + 1)}}
?
sin^2(\theta) + cos^2(\theta) = 1 so...
well i know it's sec^2(\theta) from solving the previous problem
\int...
what would be the appropriate trigonometic substition for the following:
1.) \int(7x^2-7)^{3/2}) dx
2.) \int \frac{x^2}{\sqrt{(7x^2 + 8)}}
ok i have a book with a list of trigonometric functions in front of me, but i don't get it. i don't see anything that i can sub in for.
ok i...
A major leaguer hits a baseball so that it leaves the bat at a speed of 31.4 m/s and at an angle of 38.1 above the horizontal. You can ignore air resistance
#1: At what two times is the baseball at a height of 10.9s above the point at which it left the bat? Give your answers in ascending...
yea i did use that equation,
-1.9 = v(0)(0.42) - 4.9(0.42)^2
solved for v and plugged it into
vf^2 - v(0)^2 = 2gh
dont know why i get a different answer
i even used your formula and get -2.764
coconut, i have the same problem as you(with different numbers), are you doing your homework on mp also?
-1.9 = v(0)(0.42) - 4.9(0.42)^2 after solving this for v(0) i get -2.46m/s
(-2.46)^2 =2(-9.8)h
h = -.308m
does that look right? seems incorrect to me
why did you remove cos^2(4x)?
i don't get it
for example:
= \int (1-cos^2(4x)) * cos^{10}(4x) * sin(4x) dx
you use 1 to multiply cos^{10}(4x) * sin(4x) dx then you use
cos^2(4x) to multiply cos^{10}(4x) * sin(4x) dx
right? that's how i got cos^{12}(4x)cos^2(4x)sin(4x)dx... am i...
\int sin^3(4x)cos^{10}(4x) dx
ok i know that i need to borrow the a sin, cause the sin has an odd power
\int sin^2(4x)cos^{10}(4x) sin(4x)dx
ok used one of the trig ids. on sin^2(4x)
= \int (1-cos^2(4x)) * cos^{10}(4x) * sin(4x) dx
=\int cos^{10}(4x)*sin(4x)dx - \int...
what is -ve?
and why is the answer for problem A.) a positive answer? after inserting -6.35, i get a negative answer. so... why is the answer positive instead of negative?
"For the last part, don't forget that you begin with a velocity of -6.35 m/s"
are you talking about problem C.)...
A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 6.35m/s . Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand.
A.) What is its speed after falling for a...
ok i know how you got \int{\frac{t^{15}}{15} \frac{dt}{t}} = \int{\frac{t^{14}}{15}dt
but how did you get
\int{\frac{t^{14}}{15}dt = \frac{t^{15}}{225}
i know you got the anti-derv. of it, and i tried to look in my book for a form like that, but i can't find it. can you explain to me...
Inserting the formulas you found for xman(t) and xbus(t)into the condition xman(t catch) = xbus(t catch), you obtain the following:
-b + ctcatch=1/2at^2catch
Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the...
your right,thanks.
i have another question
what is Xbus(t) the position of the bus as a function of time?
using the formula x(t) = x(0) +v(0)t + (1/2)at^2
x(t) = b +v(0)t +(1/2)ct^2
hmm what would be v(0)t? and is my setup correct? and what should i do next?
A man is running at speed c (much less than the speed of light) to catch a bus already at a stop. At t=0, when he is a distance b from the door to the bus, the bus starts moving with the positive acceleration a.
Use a coordinate system with x=0 at the door of the stopped bus
i have two...
Evaluate \int t^{14} ln(t) dt
using integration by parts and formula uv-\intv du
u=ln(t)
du=1/t
dv=t^14
v=t^15/15
lnt*t^15/15-\intt^15/15*1/t
lnt*t^15/15-t^14*log(t)
i don't know what I am doing wrong, please help
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.
Question A.) What is the...
If r^{\rightarrow} = bt^2i+ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 45.0^o with the x- and y-axes?
i & j are vectors.
i have no clue on how to start it, can anyone give me a hint?
i got x=.0610687 when trying to find the volume. so shouldn't it be x=6.11\times 10^{5}m^3=V instead of x=6.11\times 10^{-5}m^3=V?
cause x=6.11\times 10^{-5}m^3=V =.0000611 right?
so...V=.0610687 = L^3
L=\sqrt[3]{.0610687}
L = .3937974428 m^3 <-- that doesn't seem to convert to around...
let me show you what i done.
cross multiply, so i get 7.86000kgx=0.480kg
x=.0610687 <-- is that the volume?
that doesn't look right, i know I am doing it incorrectly. why did you convert 480g to .480? and what should i do with 1m^3? am i suppose to do unit conversions first? I am sorry, I...
ok I am still confuse, but let me try. ok for the first question, V=L^3 right? so...
V= 7.86 X 10^3kg and L isn't defined? where does the 480 g come in?
Iron has a property such that a 1.00m^3 volume has a mass of 7.86 X 10^3kg(density equals 7.86 X 10^3kg/m^3). You want to manufacture iron into cubes and spheres.
1.) Find the length of the side of a cube of iron that has a mass of 480 g.
2.) Find the radius of a solid sphere of iron...