Mgraph

y=-x^4+6

14 years ago

Let the distance to school S, my speed=S/30 the speed of my brother=S/40 and I catch up with him after t min. Then S/40*(t+8)=S/30*t 30t+240=40t

14 years ago

1/(x^2-90^2)

14 years ago

This value = H'(1)=-1/(2sqrt(6))

14 years ago

Critical number x=0 Increasing on interval (-Inf,0) Decreasing on interval (0,Inf) Local maxima x=0 Local minima x=none Concave up on (-Inf,-1)U(1,Inf) Concave down on (-1,1) Inflection points x=-1,x=1 Horizontal asymptotes y=0 Vertical asymptotes x=none

14 years ago

f'(x)=(x-5)^3+(x-2)3(x-5)^2=(4x-11)(x-5)^2 (A) Absmax=f(1)=64+11=75 Absmin=f(11/4)=-3^7/2^8+11=-2187/256+ +11=629/256 (B) Absmax=f(8)=173 Absmin=f(11/4)=629/256 (C) Absmax=f(9)=459 Absmin=f(4)=9

14 years ago

cos(a)cos(b)-sin(a)sin(b)=cos(a+b) Answer:cos(45+120)=cos165

14 years ago

We find a from a system: a +a+99d =7099.5 a+100d+a+199d=20799 We subtract from the 2nd the 1st: 200d=13699.5 d=68.4975 2a+99x68.4975=7099.5 a=159.12375 (0.12375=99/800)

14 years ago

p'=1/x' x'=3p^2-6p

14 years ago

m(t)=9x2^(-t/192) 192=24x8 2.8=9x2^(-t/192) t=192xln(9/2.8)/ln2 t=323.4236495

14 years ago

37 1 80 26 53

14 years ago

I think f(x)=2cos3x. f'(x)=-6sin3x maxf'(x)=6 if sin3x=-1 when x=-Pi/6 f(-Pi/6)=0 The equation of tangent y=6x+Pi

14 years ago

1)r=3cos(theta) r^2=3rcos(theta) x^2+y^2=3x 2)r=4cos(theta)-4sin(theta) r^2=4rcos(theta)-4rsin(theta) x^2+y^2=4x-4y x^2-4x +y^2+4y =0 x^2-4x+4+y^2+4y+4=4+4 (x-2)^2 +(y+2)^2=8 1)x^2+(y-1)^2=1 x^2+y^2-2y+1=1 x^2+y^2=2y r^2=2rsin(theta) r=2sin(theta) 2)(x-1)...

14 years ago

f(0)=2x0=0=0 mod 5 f(1)=2x1=2=2 mod 5 f(2)=2x2=4=4 mod 5 f(3)=2x3=6=1 mod 5 f(4)=2x4=8=3 mod 5, so f(A)={0,1,2,3,4} a/- yes b/- yes

14 years ago

30C5=30!/(5!x25!)=142,506

14 years ago

1)40C6=40!/(6!x34!)=3,838,380 2)15C3=15!/(3!x12!)=455

14 years ago

No. Sn=8(1-(-1)^9)/(1-(-1))=8*2/2=8

14 years ago

A)A=Пr^2 B)dA/dt=dA/dr*dr/dt=2Пr*dr/dt C)dA/dt=2П*7*2cm^2/s=87.96cm^2/s

14 years ago

y=... or r=...?

14 years ago

Read the terms of your problem!

14 years ago

If r=tan(theta)sec(theta) then r(cos(theta)^2)=sin(theta) (r*cos(theta))^2=r*sin(theta) x^2 = y

14 years ago

Then all right sides multiply by 5: x^2=5y or y=0.2x^2

14 years ago

Multiply both sides by r^2: r^4=r^2(2cos^2(theta)+3sin^2(theta)) (r^2)^2=2(r*cos(theta))^2+3(r*sin(theta)^2 (x^2+y^2)^2=2x^2+3y^2

14 years ago

For 0<x<2, y<0. Correct answer is (1,-0.4)

14 years ago

In decimal fraction 3.01208000 1st and 2nd zeros are significant, 3rd, 4th, 5th aren't

14 years ago

A lateral area=Pi*R*l, where R-radius, l-slant height Pi*R*10=60*Pi R=6 The volume of sphere=(4/3)*Pi*R^3= 1.33*3.14*6^3=902.06 more accurately 1.33333*3.14159*216=904.78

14 years ago

Let a-the side of the square, then sqrt(a^2+a^2)=a*sqrt(2)-diagonal. On the condition of the problem 4a=2a+a*sqrt(2)+3 a(2-sqrt(2))=3 a=3/(2-sqrt(2))=3(2+sqrt(2))/2 The perimeter of the square=4a=6(2+sqrt(2))=20.49cm

14 years ago

h'(x)=5*x^4*f(x)+x^5*f'(x) h'(-1)=5*(-1)^4*f(-1)+(-1)^5*f'(-1)= 5*1*3+(-1)*6=15-6=9

14 years ago

check condition of problem

14 years ago

Maybe we must find the square of sqrt7(cosPi/12+isinPi/12)? sqrt7=2.646 cosPi/12=0.966 sinPi/12=0.259 2.646(0.966+i0.259)=2.556+0.685i

14 years ago

The equation of the tangent to the graph y=f(x) at x=a: y=f(a)+f'(a)(x-a) (4e^x)'=4e^x y=4e^2+4e^2(x-2) y=4e^2*x-4e^2

14 years ago

Find x where 2sin(x)=3cos(x) (div by cos) 2tan(x)=3 tan(x)=1.5 x=56.31degr Area(from x=0 to x=56.31)= (3sin(56.31)+2cos(56.31)-(3sin(0)+2cos(0)) =3.60555-2=1.60555 Area(from x=56.31 to x=72)= (-2cos(72)-3sin(72))-(-2cos(56.31)-3sin(56.31))=-3.47120-(-3.60...

14 years ago

Find the point of intersection of the graph of the function and the x-axis: X^2+6X-8=0 X1=-3-sqrt(17), X2=-3+sqrt(17) CHECK FUNCTION!

14 years ago

missing data

14 years ago

Radius=2 Interval of convergence: -1<x<=3

14 years ago

f(x)=cos(Pi*x/30) is the even function. f(x+15)=cos(Pi*x/30+Pi/2)=-sin(Pi*x/30) is odd.

14 years ago

Separate the variables: 27sqrt(y)dy=2(x-3)sqrt(x^2-6x+23)dx Let z=x^2-6x+23, then dz=(x^2-6x+23)'dx= (2x-6)dx=2(x-3)dx 27sqrt(y)dy=sqrt(z)dz Integrating both sides gives 18y*sqrt(y)=(2/3)*z*sqrt(z)+C 18y*sqrt(y)=(2/3)*(x^2-6x+23)*sqrt(x^2-6x+23)+C

14 years ago

Note that sin(5pi/6)=sin(pi/6)=1/2. Area=Integral(from pi/6 to 5pi/6)(sinx-1/2)dx=(-cosx-x/2)I(pi/6 to 5pi/6)= (-cos(5pi/6)-5pi/12)-(-cos(pi/6)-pi/12)= =sqrt(3)-pi/3=0.685

14 years ago

f(x)>0 on(0,2) and f(0)=f(2)=0 The primitive of f(x) is -(4-x^2)sqrt(4-x^2)/3 The area=-(4-2^2)sqrt(4-2^2)/3+(4-0^2)sqrt(4-0^)/3=8/3

14 years ago

(0,9)

14 years ago

That approximation is the intersection of the tangent and X-axis. The equation of the tangent: y-2=5(x-3) If y=0 then x=13/5

14 years ago

(!!!)L>=0 integer. 1)Q=0 if L=-0.41 or L=0 or L=26.77 Q>=0 if 0<=L<=26 2)Q'=12+58L-3.3L^2=0 if L=17.78 Q(17)=3180.7 and Q(18)=3196.8, therefore L=18 3)The revenue R=(12L+29L^2-1.1L^3)x125-7000L R=-5500L+3625L^2-137.5L^3 R'=-5500+7250L-412.5L^2=0 if L=16.7...

14 years ago

The volume=Pi*Integral(from 0 to 1) (2y^2)^2*dy=Pi(4/5)y^5(0 to 1)=4Pi/5

14 years ago

We can find this equation using different methods (Least square method, Lagrange interpolation polynomial method, ...) Which you are applying?

14 years ago

We solve the inequality: Ix^3-3x+7-9I<e I(x-2)(x^2+2x+1)I<e I(x-2)((x-2)^2+6(x-2)+9)I<e Let e^2/100 +6e/10<1 If Ix-2I<e/10, then I(x-2)((x-2)^2+6(x-2)+9)I<= Ix-2I(Ix-2I^2+6Ix-2I+9)< (e/10)(e^2/100+6e/10+9)<(e/10)(1+9)=e, so we can take delta=epsilon/10

14 years ago

I think that the volume=pi*Int(from 0 to 1) (sqrt(x))^2dx-pi*Int(0 to 1)(x^2)^2dx= pi(1/2-1/5)

14 years ago

f(x)=(x-3)(x^2+3x+9)/x(x-3)= (x^2+3x+9)/x=x+3+9/x If x->inf then 9/x->0

14 years ago

The primitive of arcsin(x) is (x)arcsin(x)+sqrt(1-x^2). arcsin(0)=0, arcsin(1)=Pi/2.

14 years ago

(e^(t^2))'=e^(t^2)2t (e^(-t^3))'=e(-t^3)(-3t^2) velocity vector at time t=3 is (6e^9,-27e^(-27))

14 years ago

F=ma where a is the acceleration of the particle P. a=dv/dt where v is the velocity. dv/dt=(dv/dx)(dx/dt)=(dv/dx)v (dv/dx)v=36/(x^3)-9/(x^2) (v)dv=(36/(x^3)-9/(x^2))dx Integrating we get v^2/2=-18/(x^2)+9/x+C If t=0 then x=4 and v=0.5 C=-1 v^2/2=(6-x)(x-3...

14 years ago

300=100*(1.1)^t (1.1)^t=3 t=ln3/ln(1.1)=11.53 years

14 years ago

6*pi

14 years ago

Let z=1+3t^5 then dz=15t^4dt -->t^4dt=(1/15)dz The integral of (z^20)(1/15)dz=z^21/315+C= (1+3t^5)^21/315+C

14 years ago

d+q=100 10d+25q=1405 d=73, q=27

14 years ago

It is not true if (for example) A=3pi/4

14 years ago

tan(3pi/4)=cot(3pi/4)=-1 sec^2(3pi/4)=csc^2(3pi/4)=2 sqrt(2+2)=-1-1 ?

14 years ago

In the problem we must add 0<A<pi/2 that's all

14 years ago

Solve the system: n+d+q=46 n=q+11 10d=25q-100 From the 3rd -->d=2.5q-10 The 1st: (q+11)+(2.5q-10)+q=46 q=10 n=21 d=15

14 years ago

112.94cm

14 years ago

Let the first integer is (x-1) then the 2nd is x, and the 3rd is (x+1) The product is (x-1)x(x+1)=(x^2-1)x= x^3-x. If x=13 then x^3-x=13^3-13= 2197-13=2184. (12^3=1728)

14 years ago

If he uses all 3 inventions he spends (1-0.30)(1-0.45)(1-0.25)100%=28.875% of fuel. Therefore, 100-28.875=71.125% he can save.

14 years ago

E=-20cos(pi*a/4)

14 years ago

sin(x-y)=sin(x)cos(y)-sin(y)cos(x) Let y=pi/2 sin(x-pi/2)=sin(x)*0-1*cos(x)=-cos(x)

14 years ago

tan(x+pi)=tan(x) sin(pi-x)=sin(x) cos(pi/2-x)=sin(x) cot(x)tan(x)=1 1-sin^2(x)=cos^2(x)

14 years ago

The opposite event is that 4 people have birthdays on different days of the week. Its probability=(7*6*5*4)/(7^4)=120/343 Therefore, P=1-120/343

14 years ago

1)y=0, x<0 (the ray) 2)r^2=6r*cos(teta) x^2+y^2=6x x^2-6x+9-9+y^2=0 (x-3)^2+y^2=3^2 (the circle) 3)2cos(teta)=1 2r*cos(teta)=r 2x=sqrt(x^2+y^2) (x>0) 4x^2=x^2+y^2 y^2=3x^2 y=(+-)sqrt(3)x (two rays)

14 years ago

This will be a circle with a circumference 2*pi*r=240-->r=120/pi (max)A=pi*r^2=14400/pi square feet

14 years ago

We solve the system R+H+C=100 5R+3H+(1/3)C=100, where R,H,C are natural. R=100-H-C 15R+9H+C=300 15(100-H-C)+9H+C=300 6H+14C=1200 3H+7C=600, H+C<100 H=18, C=78--> R=4 H=11, C=81--> R=8 H= 4, C=84--> R=12

14 years ago

D)Sometimes. Only if a rectangle is a square.

14 years ago

b) -2y

14 years ago

f(x,y)=(x^2)(e^2x)lny f’x=((x^2)(e^2x))’lny= ((x^2)’(e^2x)+x^2(e^2x)’)lny=(2xe^2x+2x^2e^2x)lny=2xe^2x(1+x)lny f’y=(x^2e^2x)(lny)’=x^2e^2x/y

14 years ago

(1/2+1/2)^6=1^6

14 years ago

a)26C5/52C5 b)13C5/52C5 c)48/52C5 52C5=2598960 26C5=65780 13C5=1287 48=48

14 years ago

2753.963261

14 years ago

Accumulation value of an annuity (due) Sn=P*((1+i)^n-1)/d where P-the size of payments, n=17*2=34, i=0.066/2=0.033, d=i/(i+1)=0.032 140000=P*63.1024733 P=2218.61

14 years ago

5) (x^3+y^3-6)'=3x^2+3y^2*y'=0 y'=-x^2/y^2, where y=(6-x^3)^(1/3)

14 years ago

Accumulated value after 7*12=84th payment 400((1+0.006)^84-1)/0.006=43522.55653 This amount after 25 years-->158445 Let P new payments then 400000-158445=P*((1+0.006)^216-1)/0.006 241555=P*440.087 P=$549

14 years ago

Let P(n) is the price after n years, then P(n)=28000*0.92^n P(5)=28000*0.92^5=18454

14 years ago

The present value of the annuity (due) An=P*(1-v^n)/(1-v) where v=1/(1+i) P=18000, n=8, i=0.069 A8=115346

14 years ago

The selling price= the present value of the annuity due. A(42)=300*(1-v^42)/(1-v) where v=1/(1+i) i=5%/12

14 years ago

(a)The accumulated value= 1100(1+0.075/12)^144+ 100*((1+0.075/12)^144-1)/(0.075/12)=$25942 (b)1100+100*144=15500 (c)The present value of the annuity= 25942=W*(1-(1+0.075/12)^(-60))/(0.075/12) where W-the amount of each withdrawal W=519.82 (d)519.82*60=311...

14 years ago

((e^(-x^2))'=-2xe^(-x^2) ((e^(-x^2))''=-2e^(-x^2)+4x^2e^(-x^2)= 2e^(-x^2)(2x^2-1)=0 Two points: (sqrt(1/2),e^(-1/2)) (-sqrt(1/2),e^(-1/2))

14 years ago

Solve the system 10D+5N=855 30D+5(N-2)=2345

14 years ago

Solve the equation x^2-18x+81=16 x^2-18x+65=0

14 years ago

(3x+3)(-4x^2+4x-2)= 3x(-4x^2)+3(-4x^2)+3x(4x)+3(4x)+3x(-2)+ +3(-2)=

14 years ago

x/3=sin(t), y/2=cos(t) x^2/3^2=sin^2(t) + y^2/2^2=cos^2(t) _____________________ x^2/3^2+y^2/2^2=1

14 years ago

f(x)=2(x^2-6x+4)=2(x^2-6x+9-5)= =2((x-3)^2-5)=2(x-3)^2-10 The vertical axis x=3

14 years ago

From the sine formula sin(A)/a=sin(B)/b sin(B)=2.6sin(29)=1.26>1 0 solutions

14 years ago

See April 26, Daphine

14 years ago

(-1/3)cot(3x)+C

14 years ago

For simple interest 2P=P+0.14tP 0,14t=1 t= approx 7 yrs For compound interest 2P=P(1.14)^t 1.14^t=2 t=ln(2)/ln(1.14)=5.29 approx 5 yrs

14 years ago

The equation x^3-4x^2+16=0 has 1 real root x=-1.68 (approx) (x^4-4x^3+16x)'=4x^3-12x^2+16=4(x^3-3x^2+4)=4(x+1)(x^2-4x+4)=4(x+1)(x-2)^2 (x^4-4x^3+16x)''=12x^2-24x=12x(x-2)

14 years ago

sec(2x) or (sec(x))^2 ?

14 years ago

What question ?

14 years ago

f'(x)=e^(sinx)cosx*cosx+e^(sinx)(-sinx)

14 years ago

f(x)=sin(2x)/(cos2x)^2 F(x)=sec(2x)/2+C

14 years ago

I do not know how too if only angle given

14 years ago

Area=Pi*r^2 where r^2=30^2-24^2 Draw an axial section of the sphere perpendicular to the plane

14 years ago

Find the probability of the opposite event

14 years ago

If AD is bisector then angle BAC=30 AC=23*cos30=19.92 CD=AC*tan15=5.34

14 years ago