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October 22, 2014

Posts by Mgraph


Total # Posts: 309

Math
127=7^3-6^3 7^3 all 3-digit numbers, formed using 1,2,3,4,5,6,7 6^3 ... using 1,3,4,5,6,7
May 7, 2011

Math
88/4=154/7=132/6=66/3= 44 /2
May 7, 2011

Math
It depends on i- interest rate
May 7, 2011

calculus
6ln(ln(2x))+C
May 6, 2011

MATH
The product of five-digits number by four equal five-digit number-->A=1 or A=2. The last digit of number 4E is A-->A=2,E=8 4*21978=87912
May 6, 2011

Math
a=d/2-2 b=d/2+2 c=d/4 --> a+b+c+d=2d+d/4=45-->d=20 a+b+c-d=d/4=5
May 6, 2011

Math
1st glass: 60g - 20g syrup 2nd glass:120g - 30g syrup ---------------------------- 1st + 2nd:180g - 50g syrup Ans. 5/18
May 6, 2011

Algebra
Angle EOD=360/5=72 Angle EOC=2*72=144 OCE=(180-144)/2=18
May 6, 2011

MATH!
Yes
May 6, 2011

Math / Calculus
2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1) at the point (0, 0.5) 0+y'=2*2*0.25*(4*0.5*y'-1) y'=2y'-1 y'=1 Y=0.5+1*(X-0) Y=X+0.5
May 5, 2011

trig
If AD is bisector then angle BAC=30 AC=23*cos30=19.92 CD=AC*tan15=5.34
May 5, 2011

trig
I do not know how too if only angle given
May 5, 2011

Geometry
mm^2
May 5, 2011

Geometry
Area=Pi*r^2 where r^2=30^2-24^2 Draw an axial section of the sphere perpendicular to the plane
May 5, 2011

Math
Find the probability of the opposite event
May 5, 2011

calculus
f'(x)=e^(sinx)cosx*cosx+e^(sinx)(-sinx)
May 4, 2011

calculus
What question ?
May 4, 2011

calculus
f(x)=sin(2x)/(cos2x)^2 F(x)=sec(2x)/2+C
May 4, 2011

calculus
sec(2x) or (sec(x))^2 ?
May 4, 2011

math
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)
May 4, 2011

calculus
(-1/3)cot(3x)+C
May 4, 2011

math- check my answer
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
May 4, 2011

MATH AND PHYSICS HELP HELP HELP HELP HELP
See April 26, Daphine
May 4, 2011

math
From the sine formula sin(A)/a=sin(B)/b sin(B)=2.6sin(29)=1.26>1 0 solutions
May 3, 2011

math
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
May 3, 2011

Calc
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
May 3, 2011

math
(3x+3)(-4x^2+4x-2)= 3x(-4x^2)+3(-4x^2)+3x(4x)+3(4x)+3x(-2)+ +3(-2)=
May 2, 2011

math
Solve the system 10D+5N=855 30D+5(N-2)=2345
May 2, 2011

math
Solve the equation x^2-18x+81=16 x^2-18x+65=0
May 2, 2011

calculus
((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))
May 1, 2011

math
(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=31189.47 Please check
April 30, 2011

math
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
April 30, 2011

Math
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
April 30, 2011

Math (difficult)
Let P(n) is the price after n years, then P(n)=28000*0.92^n P(5)=28000*0.92^5=18454
April 30, 2011

Math(Please check. Thank You)
5) (x^3+y^3-6)'=3x^2+3y^2*y'=0 y'=-x^2/y^2, where y=(6-x^3)^(1/3)
April 30, 2011

math
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
April 30, 2011

math
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
April 30, 2011

math
2753.963261
April 30, 2011

Probability
a)26C5/52C5 b)13C5/52C5 c)48/52C5 52C5=2598960 26C5=65780 13C5=1287 48=48
April 29, 2011

Math
(1/2+1/2)^6=1^6
April 29, 2011

Math(Please help)
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
April 29, 2011

Math(Please help)
b) -2y
April 29, 2011

Math
D)Sometimes. Only if a rectangle is a square.
April 29, 2011

maths
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
April 28, 2011

math
This will be a circle with a circumference 2*pi*r=240-->r=120/pi (max)A=pi*r^2=14400/pi square feet
April 28, 2011

Trigonometry
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)
April 28, 2011

Math
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
April 27, 2011

trig
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)
April 27, 2011

trig
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)
April 27, 2011

trig
E=-20cos(pi*a/4)
April 27, 2011

college albegra
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.
April 26, 2011

Maths
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)
April 26, 2011

Geometry (math)
112.94cm
April 26, 2011

College Algebra
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
April 26, 2011

Maths
In the problem we must add 0<A<pi/2 that's all
April 26, 2011

Maths
tan(3pi/4)=cot(3pi/4)=-1 sec^2(3pi/4)=csc^2(3pi/4)=2 sqrt(2+2)=-1-1 ?
April 26, 2011

Maths
It is not true if (for example) A=3pi/4
April 26, 2011

math
d+q=100 10d+25q=1405 d=73, q=27
April 26, 2011

trig
6*pi
April 26, 2011

Calculus
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
April 26, 2011

math
300=100*(1.1)^t (1.1)^t=3 t=ln3/ln(1.1)=11.53 years
April 26, 2011

Math HELP PLEASE
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)/(x^2) therefore 3<...
April 26, 2011

calc
(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))
April 25, 2011

Calc
The primitive of arcsin(x) is (x)arcsin(x)+sqrt(1-x^2). arcsin(0)=0, arcsin(1)=Pi/2.
April 25, 2011

Math
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
April 24, 2011

calculus
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)
April 23, 2011

calculus
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
April 23, 2011

Algebra
We can find this equation using different methods (Least square method, Lagrange interpolation polynomial method, ...) Which you are applying?
April 22, 2011

Calculus
The volume=Pi*Integral(from 0 to 1) (2y^2)^2*dy=Pi(4/5)y^5(0 to 1)=4Pi/5
April 22, 2011

Calculus--Please help.
(!!!)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=16.75 ...
April 21, 2011

calculus
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
April 21, 2011

precalculus
(0,9)
April 21, 2011

calculus
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
April 21, 2011

Calculus/ trig
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
April 21, 2011

calculus
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
April 21, 2011

math
d.
April 21, 2011

precalculus
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.
April 21, 2011

Calculus ll
Radius=2 Interval of convergence: -1<x<=3
April 21, 2011

calculus
missing data
April 20, 2011

math:Calculus
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!
April 20, 2011

Calc 2: Area under the curve
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-(-3.60555)=0.13435 The total...
April 20, 2011

Calculus
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
April 19, 2011

PRECALCULUS
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
April 19, 2011

PRECALCULUS
check condition of problem
April 19, 2011

Math- Calculus
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
April 19, 2011

maths
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
April 19, 2011

Solving Geometry Problem - PLEASE HELP - Test tom!
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
April 19, 2011

Maths
In decimal fraction 3.01208000 1st and 2nd zeros are significant, 3rd, 4th, 5th aren't
April 19, 2011

Calculus
For 0<x<2, y<0. Correct answer is (1,-0.4)
April 18, 2011

trigonometry
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
April 18, 2011

Calc 2
Then all right sides multiply by 5: x^2=5y or y=0.2x^2
April 18, 2011

Calc 2
If r=tan(theta)sec(theta) then r(cos(theta)^2)=sin(theta) (r*cos(theta))^2=r*sin(theta) x^2 = y
April 18, 2011

Calc 2
Read the terms of your problem!
April 18, 2011

Calc 2
y=... or r=...?
April 18, 2011

Math
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
April 18, 2011

Algebra II
No. Sn=8(1-(-1)^9)/(1-(-1))=8*2/2=8
April 18, 2011

math!
1)40C6=40!/(6!x34!)=3,838,380 2)15C3=15!/(3!x12!)=455
April 16, 2011

math help!
30C5=30!/(5!x25!)=142,506
April 16, 2011

Discrete Math
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
April 16, 2011

PRECALCULUS HELP PLEASE!
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)^2+(y+4)^2=17 x^2-2x+1...
April 16, 2011

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