both of these exercises seem to be translating the expressions into the form aert and solving for r. a. y=3000(1.36)t=3000ert tln(1.36)=tr r=ln(1.36) y=3000etln(1.36) The other is similar as y=15etln(.29) with r=ln(.29)
both of these exercises seem to be translating the expressions into the form aert and solving for r. a. y=3000(1.36)t=3000ert tln(1.36)=tr r=ln(1.36) y=3000etln(1.36) The other is similar as y=15etln(.29) with r=ln(.29)
Also can be 4C2/52C2 as number of ways to pick 2 aces out of four/number of ways to choose 2 cards from 52. =1/221
It's a geometric series with the ratio i. The sum is (i-i24)/(1-i)=(i-1)/(1-i)=-1
Let r=length of light beam from lighthouse to spot on shore. θ=angle from perpendicular to shore from lighthouse to beam. 100/r = =cosθ or r=100secθ dr/dt = 100 secθ tanθ dθ/dt, which for θ=π/4, the right triangle is formed by 100...
∫xdx/(x2-5x+25) The derivative of the denominator is 2x-5, alas the numerator is not that. ∫xdx/(x2-5x+25)=(1/2)∫2xdx/(x2-5x+25) and now comes one of the neat secrets, add 0, as -5+5 ∫xdx/(x2-5x+25)=(1/2)∫(2x-5+5)dx/(x2-5x+25)=(1/2)∫(2x-5)dx/(x2-5x+25)+(1/2)∫5dx/(x2-5x+25) The...
A=Bert for continuous compounding A=27500, B=22000, t=4 lnA=lnB+rt r=(1/t)ln(A/B)=(1/4)ln(275/220)=0.055785888
Let D= number of ducks 2 legs per duck C=number of cows 4 legs per cow D+ C=50 2D+4C=154 now solve
y2+11y=y(y+11)=0 from which we have y=0 or y=-11
The saying is "don't look a gift horse in the teeth" - don't carefully evaluate a gift, it's a sign of ingratitude. a house of cards - a scheme or a plan which is highly unstable. a flash in the pan - not untold riches, but a few grains of gold
x - y + z = -7 x + y + z =1 x + y - z = 9 Second equation minus first equation 0+2y+0=8 y=4 Second equation minus third equation 0+0+2z=-8 z=-4 First equation plus third...
y = x - 1 and the parabola y^2 = 2x + 6 x=y+1 x=(y2-6)/2 equating the two expressions for x 2y+2=y2-6 y2-2y-8=0 (y-4)(y+2)=0 y=4, y=-2 substituting in the first equation 4=x-1 x=5 (5,4) -2=x-1 x=-1 ...
The equation is of a straight line. The two points are (8,3) and (13,7). The double point form of the straight line is (y-3)=[(7-3)/(13-8)](x-8) or y-3=[4/5](x-8) or y-3=4x/5-32/5 and so 4x/5-y=32/5-3 or 4x-5y=32-15=17 4x-5y=17 is the Ax+By=C form.
f(x)=Σx4n/(4n)! One assumes the question is based on f(x)=Σn=0∞x4n/(4n)! fiv(x)=Σn=0∞ (4n)(4n-1)(4n-2)(4n-3)x4n-4/(4n)!=Σn=4∞ x4n-4/(4n-4)! the terms with n<4 become zero. This sum is exactly f(x), as the same terms are summed in both cases. In an expanded illustration we have f(x)=1+x4/4...
ABt for A=[a] and B= [c] [b] [d] ABt=[a][c d] = [ac ad] [b] ...
4x-3y=10 -20x+15y=50 The second equation divided by -5 gives 4x-3y=-10 These are two parallel lines. The equation has no solution. The equations are inconsistent.
The fact that the arc length is 3 and he radius is 7 is a complete specification of the size of the angle which is length of arc/length of circle =angle in radians /2π angle in radians=2π×3/(2π×7)=3/7 radians = (180/π)×3/7 degrees=24.55533408 Either start at the x-axis and go up 3/7...
Let E=set of even numbered cards E=(E∩A)∪(E∩B) a disjoint union P(E)=P(E∩A)+P(E∩B)=5/10+2/5=5/10+4/10=9/10!!! P(A|E)=P(A∩E)/P(E)=(5/10)/(9/10)=5/9 Sorry for the arithmetic error. Same theorem, different statement.
-x-2y=2 4x-2y=8 add 4×first equation to second and have -x - 2y= 2 -10y=16 from which we get y=-8/5 substituting in first equation we have -x+16/5=10/5 or x=6/5 Solution = (6/5,-8/5)
t=time at reduced speed 70×(7-t)+10t=250 490-70t+10t=250 240=60t t=4 N.B. 70×3+10×4=250
If you divide the $90000 into 7+8+10=25 parts, each of 90000/25=3600, the amounts are 7×3600=25200 8×3600=28800 10×3600=36000