.5×7 + x = the acid in the final soloution 7+x is the amount of the final solution (3.5+x)/(7+x)=.7 3.5+x=.7(7+x) 3.5+x=4.9+.7x .3x=1.4 x=14/3=4 2/3 gal Note:3 1/2 + 4 2/3 = 8 1/6 total acid after addition of acid 7+4...
.5×7 + x = the acid in the final soloution 7+x is the amount of the final solution (3.5+x)/(7+x)=.7 3.5+x=.7(7+x) 3.5+x=4.9+.7x .3x=1.4 x=14/3=4 2/3 gal Note:3 1/2 + 4 2/3 = 8 1/6 total acid after addition of acid 7+4...
x6-19x3-216=0 Let y=x3 and have y2-19y-216=0 which may be factored as (y-27)(y+8)=0 which gives that y=27, y=-8 and so that x=3 or x=-2
l = long side, w = short side l+w=11 half the perimeter l2+w2=101 Pythagoras l+w=11 gives that (l+w)2=l2+2lw+w2=121 or that 2lw=20 2lw=20 or lw=10 l+w=11 lw=10...
Things equal to the same thing are equal to each other, or equality is transitive. Which, for angles says that if angle A is congruent to angle D and angle D is congruent to angle E then angle A is congruent to angle E. This is more about the nature of equality than that of...
-2x+8≤-5 subtract 8 from both sides to get -2x ≤ -13 divide both sides by -2 to get x≥13/2 6<2x multiply both sides by -1 to get -6>-2x add 8 to both sides to get 2>-2x+8 or -2x+8<2 the statement of the problem is not quite "or"...
L=length, W=width L=2W-10 the condition given L+L+W+W=400 the perimeter 2L+2W=400 L+W=200 2W-10+W=200 3W=210 W=70 L=130
The number of successes is 12C10 = 12C2 choose 10 face cards from the 12 in the deck. The number of ways to choose 10 cards from the deck is 52C10. The probability is 12C2/52C10 The calculation gives (12×11/2)/((52×...×43)/10!)=4.171927873E-09 as the probability
If this question is asked in 5th grade, its answer should be explained at that level. There are 10 teams to choose for 1st place, having done that there are 9 teams that could be in 2nd place, and that done there remain 8 teams that could be in 3rd place. The total number of choices is...
[sec(π/4)cos(2π/3)]/[tan(π/6)/csc(3π/4)]. =cos(2π/3)/[cos(π/4)tan(π/6)sin(3π/4)] =-.5/[(√2/2)(√3/3)√2/2)]=-3/√3=-√3 The only tricky parts are finding the reference angles for 2π/3 and 3π/4
D. x=sin2y dx=2sinycosydy √x/√(1-x)=siny/cosy integrand is (siny/cosy)2sinycosydy=2sin2ydy at x=0 y=0 and at x=1/2 y=π/4 ∫2sin2ydy from 0 to π/4
Original fraction 5d/d. Adjusted fraction (5d+12)/(d+12)=2 5d+12=2d+24, 3d=12, d=4 20/4 original fraction, 32/16 when 12 added to both numerator and denominator
9(log7)/(log3)=(32)log37=32log37=3log349=49
a. 4C2/52C2=1/221 two aces out of four/choose 2 cards from deck b. 48C2/52C2=188/221 two cards not aces/choose 2 cards from deck
5z-20/6z-24=5(z-4)/6(z-4)=5/6 if z≠4 and may be defined as 5/6 by limit if z=4.
Distance= Rate times time and so time=Distance/Rate time=85/√10,800 if you like rational denominators time=85√10,800 /10,800
(x-3)2=25 x-3=±5 x=3±5=8 or -2
∑n=1n=∞(n5/5n^4) The ratio test requires that we look at the ratio of the n+1st term to the nth term ((n+1)5/n5)×5n^4-(n+1)^4 ((n+1)/n)5→1 as n→∞ and for the second term 5n^4-(n+1)^4=5-(4n^3+6n^2+4n+1)→0 as n→∞ The limit for the ratio test is 0. A...
12∑1/3j for j=0 to 3
tgo=d/30 tret=d/40 ave speed=2d/(d/30+d/40)=2/(1/30+1/40) =2(1200)/70=240/7=34 2/7 mph
The quadratic equation 5x2+7x-6=0 has discriminant 72-4(5)(-6)=49+120=169, it has both its roots real. Since we are so near the solution, the quadratic formula gives x=(-7±13)/10=-2, 6/10 N.B. 5×4+7×(-2)-6=0 as does 5×(3/5)2+7×3/5-6=5×9/25+21/5-6=0