# Posts by agrin04

Total # Posts: 71

**math**

Find the area first, then multiply the result with the price

**calculus**

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)

**math**

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

**Math (Trig.)**

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 travelled will be...

**Calculus**

Oh sorry. It should be: S sin(u) du

**Calculus**

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

**maths**

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 right side. 9x - ...

**Math**

y=x/8

**calc**

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<0: d/dx (2(x^2 - 3x)) = ?

**calculus**

A = l* w dA/dt = l*dw/dt + w*dl/dt Plug in the values and calculate

**Geometry**

Diagonal = 16 = sqrt(2x^2) x = 8*sqrt(2) mm Perimeter = 4*x = ?

**maths**

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)

**maths**

And the question is?

**math**

w^(4+1+5) = w^10

**Math(Please help)**

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 the value after ...

**Math(Please help)**

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

**math algebra..help please urgent! due in 20mins**

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: Quotient = 2x^2 - 5x - 4

**math algebra..help please urgent! due in 20mins**

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 + (80/3)/(6x+4) Now, use the long...

**Word Problem**

Just add the two costs: 5x^2 + 4x - 7 + 8x + 8 =?

**Algebra**

(2x-3)(x+1)

**Pre-Calculus**

Use the cosine rule: x^2 = 52^2 + 23^2 - 2(52)(23)cos96 After calculating, you'll get x = 59

**PRE CALC**

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)) = ?

**Trigonometry**

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 quotient is the ...

**integrals**

|(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

**Math**

10 = 2 x 5 60 = 2^2 x 3 x 5 32 = 2^5 GCF = 2 LCM = 2^5 x 3 x 5 = ?

**Calculus**

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)

**math**

(a) {people who are more than 20 years old and enrolled in college} (b) {4} (c) {}

**geometry**

Right triangle

**math**

(a) 3log (3^3) = 3 (b) 5log (5^(-1)) = -1 (c) f(e^x) = ln(e^x) = x

**trigonometry**

cos ¤ = (4^2 + 5^2 - 6^2)/(2*4*5)

**Algebra**

(w^9 - 2y^5)(w^9 - 7y^5)

**trig**

Length of string = 100/cos(57)

**precalculus**

(a) 1500*2^(t/0.5) = 1500*2^(2t) (b) 1500*2^(2*20/60) = ? (c) 1500*2^(2*9) = ?

**math**

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) dx = = |(from -pi to +...

**trig**

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

**Algebra 2**

Only consider the denominator. this expression is undefined when p^2 - 49 = 0 Factorise this and you'll get the results

**calculus**

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

**Preclaculus Help Another**

cos(40)/sin(40) - sin(50)/sin(40) = sin(90-40)/sin(40) - sin(50)/sin(40) = 0

**Trig**

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

**Math**

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

**Calculus B**

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

**math**

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

**Advanced Math**

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 new equation with a ...

**Advanced Math**

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 (sqrt(2)/2)-a...

**Calculus AB**

|2^(-1/t)*(1/t^2) d(-1/t)/(1/t^2)= |2^(-1/t) d(-1/t) = 2^(-1/t)/ln2 + const

**maths**

Oops... Sorry. Something missing. Originally, there are 2 mice So, jojo will have 2048 mice after: 11 - 1 = 10 months

**maths**

2048 = 2^n n = 11

**Calc**

From point (2,1) to (1,3): u = <1-2,3-1> = <-1,2> 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 = <-1,2>/sqrt(5) <dz/dx,dz/dy>•<-1/sqrt(5),2/sqrt(5)> = -2/sqrt(5) -dz...

**Directional derivative**

df(x,y)/dx = 2x df(1,2)/dx = 2 df(x,y)/dy = -2y df(1,2)/dy = -4 Directional derivative = = <2,-4> • <(3/5),-(4/5)> = (6/5) + (16/5) = 22/5

**Mathematics**

What do you have if you squared a value or a number? You'll get a positive number the whole time. Except for 0, of course. In this case, the range will be at least 0, or >=0, regardless of the value of x

**MATH grade 12**

Just change the variable x with the value x = 2. So, we get: f(x) = 2x^3 + mx^2 + nx - 3 f(2) = 0 = 2(8) + 4m + 2n - 3 0 = 13 + 4m + 2n. ..(1) g(x) = 3mx^2 + 2nx + 4 g(2) = 0 = 12m + 4n + 4 0 = 6m + 2n + 2. ..(2) Using elimination and substitution for both equation (1) and (2...

**math**

|x^2*cos^2(x) dx = =|x^2*(1/2)(1+cos(2x)) dx =(1/2)|x^2 dx + (1/2)|x^2*cos(2x) dx =(1/6)x^3 + (1/2)|x^2*cos(2x) dx Using integration by part: u = x^2 du = 2x dx dv = cos(2x) dx v = (1/2) sin(2x) |x^2*cos(2x) dx = = (1/2)x^2*sin(2x) - |xsin(2x) dx Again, using integration by ...

**calculus**

All real numbers

**math**

|cot^4(1-2x) dx = = |cot^2(1-2x)*cot^2(1-2x) dx = |{cosec^2(1-2x) - 1}*cot^2(1-2x) dx = |cosec^2(1-2x)*cot^2(1-2x) dx - |cot^2(1-2x) dx = |cosec^2(1-2x)*cot^2(1-2x) d(cot(1-2x))/(-cosec^2(1-2x)*(-2)) - |{cosec^2(1-2x) - 1} dx = (1/2)|cot^2(1-2x) d(cot(1-2x)) - |cosec^2(1-2x) ...

**Math (Trig)**

1 rotation = 2pi 1/8 rotation = 1/8 x 2pi = pi/4 State y as the height difference created from the rotation of the wheel and the bottom of the ferris wheel (not the ground) height = y + 1 = r - r cos {(2pi/20)t - (pi/4)} +1 = r(1 - cos pi{(t/10) - (1/4)}) +1 = 10(1 - cos pi{(t...

**math**

State that the difference to be 'b' y = x + b b = y - x. ..(1) z = y + b = x + 2b b = (z - x)/2. ..(2) Combining equations (1) and (2): b = y - x = (z - x)/2 y = (z + x)/2

**Geometric**

(1) ar^(n-1) + ar^(n-2) + ar^(n-3) = 1024 (a + ar + ar^2) ar^(n-3) (1 + r + r^2) = 1024a (1 + r + r^2) r^(n-3) = 1024 It is known that the third term is 5, so: ar^2 = 5 The last term will be: ar^(n-1) = ar^2. r^(n-3) = 5 x 1024 = ? (You can finish the rest of it, right?) (2) ...

**maths**

(1) Convert into cm first, so: 420 m = ___ cm Then, divide that number with 8000 (2) Multiply 31.4 with 15000 (in cm), then convert the result into km

**Geometry**

Supplementary angles: x + y = 180 From the problem, we can state that: y = (1/4)x Substituting back to the first equation, we have: x + (1/4)x = 180 (5/4)x = 180 x = ? (You can calculate this yourself) As to find the supplementary angle, you can use the result for x by using ...

**calculus**

Find the first derivative of the function: dy/dx = (2-x)^(1/2) - (1/2)x.(2-x)^(-1/2) = (4-3x)/2sqrt(2-x) You have to be careful in this matter, especially since you've encountered a square root form in the denominator. * take the denominator, ignore the constant in front ...

**Math**

u = 20 ÷ 6 2/3 = 20 ÷ 20/3 = 20 x 3/20 = ?

**math**

Say that: dimes = d Quarters = q The algebraic expression: d = 4q - 3 Total coins = 47, so: 47 = d + q 47 = 4q - 3 + q q = 10 So, we can have d = 4q - 3 = 37

**math**

Say that the other length is expressed in y. So, the answer will be: y = (59 - x) cm

**math (calculus)**

6 sec^2 (5?/4) - 7 sin (5?/4) = = 6 (1/(-1/sqrt(2)))^2 - 7 (-1/sqrt(2)) = 6 (2) + 7/sqrt(2) = 12 + 7/sqrt(2)

**Math**

Since you cut a 2 in. x 2 in. square from each corner, you'll get a 2 in. height Length now becomes = 22 - 4 = 18 in. Width now becomes = 16 - 4 = 12 in. The volume of the box will then be: Vol = length x width x height = 18 x 12 x 2 = 432 in^3 Note: the volume is the same...

**Math**

Since you cut a 2 in. x 2 in. square from each corner, you'll get a 2 in. height The volume of the box will then be: Vol = length x width x height = 22 x 16 x 2 = 704 in^3 Note: the volume is the same for all cases, regardless of the condition of the lid (closed/open)

**math**

From the basic formula: f(x) = sin (u), where u = g(x) f'(x) = u'. cos(u) We now set that: u = e^4x u' = 4e^4x So, we have: f(x) = sin(e^4x) f'(x) = 4e^4x. cos(e^4x)

**math**

|cos^nx dx = |cos^(n-1)x. cosx dx Take: u = cos^(n-1)x du = (n-1) cos^(n-2)x. (-sinx) dx dv = cosx dx v = sinx By integration by part formula we have: |cos^nx dx = = cos^(n-1)x. sinx + (n-1)|sin^2x. cos^(n-2)x dx = cos^(n-1)x. sinx + (n-1)|(1-cos^2x). cos^(n-2)x dx = cos^(n-1)...

**math**

The change in altitude: da/dt = 3 cm/min The change in area: dA/dt = 7 cm^2/min The change in base: db/dt From the formula of area of triangle: A = (a x b)/2 66 = (22 x b)/2 b = 6 cm Differentiate the formula above with respect to time: dA/dt = (b. da/dt + a. db/dt)/2 7 = ((...

**math**

State as followings: P(P) = #students take physics = 32 P(C) = #students take chemistry = 51 P(P and C) = 15 P(neither) = 10 Total students = P(P) + P(C) - P(P and C) + P(neither) = 78

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