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a)f(x)=(x+2)(x-1)(x-2) c)From the graph a>=2, b>=0. The equation of the tangent y=b+(3a^2-2a-4)(x-a) and if x=0 y=b-(3a^3-2a^2-4a) or y=a^3-a^2-4a+4-(3a^3-2a^2-4a) y=-2a^3+a^2+4=(2-a)(6+3a+2a^2)-8 If a>2 then y
The equation of a circle with the center at origin and passing through (1,sqrt3)is x^2+y^2=2^2 The distance between the point and semicircle is the difference R1-R2=4-2=2
B)CLOCKWISE In 1st quadrant (for ex.)dx/dt>0, so x increases; C)dx/dt=d(cos(theta))/dt= d(cos(theta))/d(theta)*d(theta)/dt= -sin(theta)*d(theta)/dx, so dx/dt=-y*d(theta)/dt =>d(theta)/dt=-1
4sin^2(x)cos^2(x)=(sin2x)^2 (sin2x/cos2x)dx=(-1/2)d(cos2x)/cos2x
F(x) = f(xf(xf(x))) F'(x) = f'(xf(xf(x))) d/dx xf(xf(x)) F'(x) = f'(xf(xf(x))) [ f(xf(x)) + xf'(xf(x)) d/dx xf(x) ] F'(x) = f'(xf(xf(x))) [ f(xf(x)) + xf'(xf(x)) [f(x) + xf'(x)] ] F'(1) = f'(1*f(1*f(1))) [ f(1*f(1)) + 1*f'(1*f(1)) [f(1) + 1*f'(1)] ] F'(1)
-52 mod 13 = 0 -45-(-52)=7
3,4,5,6 2,3,6,7 2,4,5,7 1,4,5,8 1,2,7,8 1,3,6,8 1,4,5,8 (4,4,5,5)?
sqrt-2/3 ?
We integrate by parts twice: 1)u=x^2, du=2xdx, dv=cos(x/4)dx, v=4sin(x/4) 2)u=8x, du=8dx, dv=sin(x/4)dx, v=-4cos(x/4) The answer I have already sent.
sqrt=square root?
You're welcome.
AB=sqrt(((-5)-(-4))^2+(1-(-3))^2)= sqrt((-1)^2+4^2)=sqrt(17) The perimeter=4*AB
3x^2-15x=6x^2-16x 3x^2-x=0 x(3x-1)=0 x=0 or x=1/3
Let it takes Todd X hours; Tom Y hours. X=Y+3 For one hour, Todd paints 1/X part of the bedroom, Tom - 1/Y; together - 1/X+1/Y 1/(1/X+1/Y)=2 2(X+Y)=XY 2(2Y+3)=(Y+3)Y 4Y+6=Y^2+3Y Y^2-Y-6=0 => Y=3, X=6
B3=20 The sum=B1/(1-q)=3*B1 =>q=2/3 B3=B1*q^2 20=B1*4/9 =>B1=45
3sin^2(x)+sin(x)cos(x)=2sin^2(x)+2cos^2(x) sin^2(x)+sin(x)cos(x)-2cos^2(x)=0 Divide by cos^2(x) tan^2(x)+tan(x)-2=0 (tan(x)+2)(tan(x)-1)=0 tan(x)=-2 => x=-Arctan(2)+pi*n tan(x)=1 => x=pi/4+pi*n
The average velocity=distance/time, time=60s distance=2*10^2/2+20*45+4*5^2/2=1050m
Lengths of the sides of these squares are equal to y. The volume=Int(from -4 to 4)y^2dx= 2*Int(from 0 to 4)(16-x^2)dx= 2*(16x-x^3/3)(from 0 to 4)= 2(64-64/3)
We want to "guess" that limh(x)=1/3 if x->0 h(1)=(tan(1)-1)/1^3=0.557408 h(0.1)=0.334672 h(0.005)=(tan(0.005)-0.005)/0.005^3=0.333342
f'(x)=11x^10*h(x)+x^11*h'(x) f'(-1)=11(-1)^10*h(-1)+(-1)^11*h'(-1)
Antiderivative (one of) of y is 4sin(x/4)(x^2-32)+32x*cos(x/4)
tan(alpha)=0.590720839 alpha=tan^-1(0.590720839)=0.630778859rad= 36 degrees
124-6*(0+15)=34
I agree with bobpursley if 0
What is f(4)?
Straight line is a special case of the curve just like the natural numbers are a special case of integers
f(0)=(0-1/2)^2(0+1)
There were 3 eggs.
cos^2(t)=(1-x)/3 sin^2(t)=(y+2)/4 Add two equations 1=(1-x)/3+(y+2)/4 12=4-4x+3y+6 y=(4/3)x+2/3
Let R(x1,y1), S(x2,y2), then x1=(2*2+8)/3=4, y1=(2*3+(-9))/3=-1 x2=(2+2*8)/3=6, y2=(3+2*(-9))/3=-5
x=7cos(30), y=7sin(30)
(x-4)^2+y^2=16 x^2-8x+16+y^2=16 x^2+y^2=8x r^2=8r*cos(theta) r=8cos(theta)
For all x: 0
y-2=4x-x^2 The volume=pi*Int.[0,4](4x-x^2)^2dx= pi*Int.[0,4](16x^2-8x^3+x^4)dx= pi*(16/3x^3-2x^4+1/5x^5)[0,4]= pi*(1024/3-512+1024/5)=pi*1024/30
First find (x-3i)(x+3i)=x^2+9 and then (x-3)(x^2+9)=x^3-3x^2+9x-27
AC=BD...
Int=1/5(sin(t)-t*COS(t))+C
t=x^5, dt=5x^4dx x^9*sin(x^5)dx=x^5*sin(x^5)*x^4*dx= t*sin(t)*(1/5)dt Int=1/5(sin(t)-t*sin(t))+C, t=x^5
p=.001, n=2000 a)Expected number of errors=np=2 b)P(k)=2000Ck*.001^k*.999^(2000-k) c)P(0)=.999^2000 d)P(1)=2000*.001*.999^1999 e)P(k>=1)=1-P(0)
Let we invest $X in 1st fund, then 0.115X+0.14(15000-X)=1950 0.115X+2100-0.14X=1950 150=0.025X X=6000
34=2x-10
y(12-x-y)=4(12-4-4)=16
If z=cos(2q) then dz=-2sin(2q)dq. Integral=-1/2*Integral z^4dz= -1/10*z^5+C
f'(c)=(b^2-a^2)/(b-a)=b+a=2c=>c=(a+b)/2
66-5*3-5*9
Reiny found a positive root but there is a negative root x=-0.55 (approximately) We can started with x=-1
a^2-b^2=(a-b)(a+b) 4-x^2=2^2-x^2
1)x^2(x^2-9)=x^2(x-3)(x+3)>0 x>3 or x
17-10
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