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agrin04
Answers (70)
dy/dx = y cos(x) dy/y = cos(x) dx ln y + const = sin(x) + const ln y = sin(x) + const ln 3 = sin(0) + const const = ln 3 ln y = sin(x) + ln 3 ln y - ln 3 = sin(x) ln (y/3) = sin(x) y/3 = e^sin(x) y = 3e^sin(x)
Find the area first, then multiply the result with the price
Assume that the perimeter is x x = 28 + (4/9)x + (2/5)x x(1 - 4/9 - 2/5) = 28 Solve the calculation above, and you'll get the answer
Assume that the distance travelled before turning 60 degrees is x The total distance travelled = x+10 Distance A-B in straight line = x + 6 Use the cosine rule: (x+6)^2 = x^2 + 100 - (2*x*10*cos(120)) Finish the calculation, you'll find x. Total distance
Oh sorry. It should be: S sin(u) du
The first problem is correct. As for the second problem, take: u = ln x du = (1/x) dx So: S (ln u) du = ? Try it yourself
Cross multiple means: nominator of left side x denominator of right side = nominator of right side x denominator of left side So for this case: 9 * (x-3) = 4 * (x-7) 9x - 27 = 4x - 28 Now, let's collect the variable x on the left side, and the rest on the
y=x/8
For |x|, we have 2 situations. |x| = x, for x >= 0 |x| = -x, for x < 0 So: ¤ for x>=0: d/dx (2(x^2 + 3x)) = ? ¤ for x
A = l* w dA/dt = l*dw/dt + w*dl/dt Plug in the values and calculate
Diagonal = 16 = sqrt(2x^2) x = 8*sqrt(2) mm Perimeter = 4*x = ?
And the question is?
Each monkey has to eat one banana? So the total will be 8 bananas? The answer will be just 1 minute (assuming that all monkeys eat at the same time)
If you are not given the value, that means you have to write the steps I told you until the very last equation. If you are given the value (of r), you don't have to put the value in every equation. Just follow my steps until the last equation, then plug
Yes
w^(4+1+5) = w^10
S = 4pi*r^2 dS/dr = 4pi*2r = 8pi*r V = (4/3)pi*r^3 dV/dr = 4pi*r^2 dV/dt = dV/dr * dr/dt 2 = 4pi*r^2 * dr/dt dr/dt = 1/(2pi*r^2) dS/dt = dS/dr * dr/dt = 8pi*r * 1/(2pi*r^2) = 4/r If r = 12, just plug the value to the last equation
Oops... Sorry. I've made a few errors in my calculation. Should be: -2/3 | 12 -22 -44 -16 | -8 20 16 ---------------- 12 -30 -24 0 Since the denominator has 1 degree of polynomial and its coefficient is 6, we divide the quotient with 6. So, we have:
You can use either long division method or Horner. -2/3 | 12 -22 -44 -16 | -8 -20 128/3 --------------------- 12 -30 -64 80/3 Quotient = 12x^2 - 30x - 64 Remainder = 80/3 Or, you can write as: 12x^3 - 22x^2 - 44x - 16 / 6x+4 = 12x^2 - 30x - 64 +
Just add the two costs: 5x^2 + 4x - 7 + 8x + 8 =?
|(x+6)/5 dx = = (1/5)|(x+6) dx = (1/5)*{(1/2)x^2 + 6x} + const = (1/10)x^2 + (6/5)x + const
10 = 2 x 5 60 = 2^2 x 3 x 5 32 = 2^5 GCF = 2 LCM = 2^5 x 3 x 5 = ?
For logarithmic expression, the value of u in ln u should be more than 0. So: (13x + 6)/(5 - 17x) > 0 Use the number line to solve this, you'll get the result. If my calculations are correct, then: (-6/13,5/17)
Use the cosine rule: x^2 = 52^2 + 23^2 - 2(52)(23)cos96 After calculating, you'll get x = 59
Vertical asymptote: denominator = 0, so: x^2 - 9 = 0 Horizontal: lim x->(infinity) f(x) Since the degree of nominator is higher than the denominator, then the horizontal asymptote does not exist. Slant: use long division method to find the quotient. That
(2x-3)(x+1)
1. tan(36 - 2) = tan 34 2. sin ? = 4/5 --> cos ? = 3/5 cos ? = -9/41 --> sin ? = -40/41 sin(?-?) = sin?*cos? - cos?*sin? = ? 3. tan x = sqrt(143) tan 2x = 2tanx/(1 - tan^2(x)) = ?
(a) {people who are more than 20 years old and enrolled in college} (b) {4} (c) {}
Right triangle
(a) 3log (3^3) = 3 (b) 5log (5^(-1)) = -1 (c) f(e^x) = ln(e^x) = x
(w^9 - 2y^5)(w^9 - 7y^5)
Length of string = 100/cos(57)
cos ¤ = (4^2 + 5^2 - 6^2)/(2*4*5)
(a) 1500*2^(t/0.5) = 1500*2^(2t) (b) 1500*2^(2*20/60) = ? (c) 1500*2^(2*9) = ?
Assume that for all cases that both m and n are integers. For m = n: |(from -pi to +pi) cos^2(mx) dx = = |(from -pi to +pi) (1/2)(1 + cos(2mx)) dx = (1/2)(pi + pi) + (1/4){sin(2mpi) - sin(-2mpi)} = pi For m not equal n: |(from -pi to +pi) cos(mx) cos(nx)
sec^4(x) - tan^4(x) = = (1 + tan^2(x))^2 - tan^4(x) = 1 + 2tan^2(x) + tan^4(x) - tan^4(x) = 1 + 2tan^2(x) QED
f(x) = 1/(6x^2) = (1/6)x^(-2) f'(x) = (-1/3)x^(-3) f''(x) = x^(-4) = 1/x^4 As to answer question b, just change x with 3 and calculate the result
Only consider the denominator. this expression is undefined when p^2 - 49 = 0 Factorise this and you'll get the results
2r - 3r sin¤ = 6 2sqrt(x^2 + y^2) - 3y = 6 2sqrt(x^2 + y^2) = 3y + 6 =3(y+2) 4(x^2 + y^2) = 9(y + 2)^2
cos(40)/sin(40) - sin(50)/sin(40) = sin(90-40)/sin(40) - sin(50)/sin(40) = 0
You can use long division method or you can use Horner method. Horner is easier and faster. 3 | 3 -11. 10 -12 | 9 -6 12 --------------- 3 -2 4 0 So, the quotient is 3x^2 - 2x + 4
Take: u = ln(2x+1) du = 2/(2x+1) dx dv = dx v = x |ln(2x+1) dx = = xln(2x+1) - |2x/(2x+1) dx = xln(2x+1) - |{1 - 1/(2x+1)} dx = xln(2x+1) - |dx + |1/(2x+1) dx = xln(2x+1) - x + (1/2)ln(2x+1) + const
Height after 4th bounce = 8x(7/8)^4 Total = 8 + 8*(7/8)*2 + 8*(7/8)^2*2 + 8*(7/8)^3*2 + 8*(7/8)^4
|2^(-1/t)*(1/t^2) d(-1/t)/(1/t^2)= |2^(-1/t) d(-1/t) = 2^(-1/t)/ln2 + const
In general: (x-a)^2 + (y-b)^2 = r^2 Center point in 8x + 5y = 8, meaning: 8a + 5b = 8. ..(1) Circle passing (2,1): (2-a)^2 + (1-b)^2 = r^2. ..(2) Circle passing (3,5): (3-a)^2 + (5-b)^2 = r^2. ..(3) Use equations (2) and (3) to eliminate r, we now have a
Circle equation generally: (x-a)^2 + (y-b)^2 = r^2 The center is in x-axis (b = 0), giving: (x-a)^2 + y^2 = 1 ((sqrt(2)/2)-a)^2 + (sqrt(2)/2)^2 = 1 ((sqrt(2)/2)-a)^2 + (1/2) = 1 (sqrt(2)/2)-a = ±sqrt(1/2) (sqrt(2)/2)-a = sqrt(1/2) a = 0 So: x^2 + y^2 = 1
Oops... Sorry. Something missing. Originally, there are 2 mice So, jojo will have 2048 mice after: 11 - 1 = 10 months
2048 = 2^n n = 11
From point (2,1) to (1,3): u = = Note that vector u above is not a unit vector, so we need to make this a unit vector by dividing it with its magnitude. u = /sqrt(5) • = -2/sqrt(5) -dz/dx + 2dz/dy = -2. ..(1) From point (2,1) to (5,5): v = = Again, this
df(x,y)/dx = 2x df(1,2)/dx = 2 df(x,y)/dy = -2y df(1,2)/dy = -4 Directional derivative = = • = (6/5) + (16/5) = 22/5
