Mgraph
This page lists questions and answers that were posted by visitors named Mgraph.
Questions
The following questions were asked by visitors named Mgraph.
Answers
The following answers were posted by visitors named 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