# Posts by steve

Total # Posts: 52,300

**Math**

#1 ok #2 h(t) = 3(t-1)^2+2 is apparently a typo, but your solution is correct as written #3 ok if you ignore 3.00-0.5(2)=$11.5 good work

**Math**

I read it as meaning that 2/5 of the original amount was spent on fruits, giving: 50 on chocolate 180 on fruit then subtract all that from the original 450.

**Algebra**

Midway between 14 and 26 is 20. |x-20| <= 6

**Algebra**

A is right. Although, it could just as easily be 57 <= t <= 73 On the number line plot the points 57 and 73, and shade the line between them. Then make sure the two points are either open or closed circles, depending on whether the endpoints are to be included.

**Algebra 2**

If you want rational coefficients, then that ?6 has to go, just like the i terms. So, you need f(x) = (x-2)(x-2i)(x+2i)(x-(4-?6))(x-(4+?6)) = (x-2)(x^2+4)((x-4)+?6)((x-4)-?6) = (x-2)(x^2+4)(x^2-8x+10)

**algebra 1**

for all the details of long division, you can verify your results by entering your polynomials at http://calc101.com/webMathematica/long-divide.jsp

**GPA**

there's a lot not being said, but if all the classes are weighted equally, they would indeed "cancel out." You don't say what an A means as far as percentages, so it's hard to say just what the effect is. If you have all those A's, including math, you...

**math**

1:2:3 = 7:14:21

**math test help**

that covers a lot of ground. Don't expect an online course here. That's what your text was for. Any general explanations we can provide will look a lot like your text, dontcha think? Or, you can google topics of interest.

**math test help**

but if you have a particular problem that is vexing you, we can help get to the solution. If you haven't studied the topic by now, it may be too late...

**math**

The sum of their ages 2 years ago was 8*20 = 160 So, if the new child's age is x, 160 + 8*2 + x = 9*20 x = 4

**math**

Hmmm. I'm not sure I agree. Since A&B can fill the tank so fast, and it takes about 9 times as long to fill it with the leak, the leak must be pretty fast in relation to the fill speeds. You sure that 5 minutes is right?

**math**

A&B together take 1/(1/5 + 1/20) = 4 minutes to fill the tank. So, if the leak takes x minutes to empty the tank, 1/4 - 1/x = 1/(30+4) x = 68/15 = 4 8/15 minutes

**calculus**

use product rule

**calculus**

these are all just powers of x. Use the power rule: d/dx x^n = n x^(n-1)

**math**

If his speed was x, then since time = distance/speed, 30/15 + 16/x = 10/3

**Math**

If the two legs are x, then x^2/2 = 25/2 x = 5 so now you can get the hypotenuse. Or, since a = s^2/2 h = s?2 h^2 = 2s^2 = 4a h = 2?a

**Maths**

a little synthetic division shows that (x-?2) does not divide the cubic. In fact f(?2) = 4-2?2 Check for typos and try again. Once you have done the division, you will be left with a quadratic, which you can then solve in the usual way.

**Science**

you have a right triangle. Shouldn't be too hard to get one of the angles, eh?

**Science**

another right triangle.

**Science**

draw the parallelogram. The two vectors have magnitude 3x and 5x, so (3x)^2 + (5x)^2 - 2(3x)(5x)cos(120°) = 35 find x from that...

**mathematics**

impatient much?

**mathematics**

average speed is totaldistance/totaltime: (100+40)/(8+2) = 14 km/hr more like a bicycle, at that speed.

**Math**

P(9,4) = 9*8*7*6 C(9,4) = P(9,4)/4!

**physics/calculus**

the product rule. Both t and y are functions of t, so d/dt (bty) = b(y dt/dt + t dy/dt) = b(y + ty')

**Math**

5P4 = 120 3*2*5

**Math**

c'mon. Just substitute values for n, from 1 to 5: a1 = 3(4*1-1) = 3(4-1) = 3(3) = 9 and so on

**Math**

If his average so far is x, then 9x+9 = 10(x-2) x=29 9*29+9 = 270

**Algebra**

if the width is w, then the length is w+7, so the diagonal is w+14: w^2 + (w+7)^2 = (w+14)^2

**maths**

well, the angle ? between the two boats' headings (not bearings!) can be found via the law of cosines: 12^2 = 8^2+10^2 - 2*8*10 cos?

**Math**

how can angle BAC be measured in cm?

**Math (Calc) (Differential Equation Solution)**

huh? what's to start? The 1st equation is the solution to the second. All you have to do is find C such that y(x^2+y) = C goes through (0,2). 2(0+2) = C

**math**

well, 184/8 = 23, so ...

**math**

If the original price was x, then 1.4x*.95 - 1.4x*.90 = 14 x = 200 1.4*200 - 1.4*200*.8 = 224 so, the profit would be 24, or 12%.

**physics**

the weight of the man is unaffected by all the horizontal pushing and shoving: mg = 60kg * 9.8 m/s^2 = 588N

**physics**

since speed = distance/time, just plug in your numbers: ((2?*0.6)*5 m)/(2s) = 3? m/s

**Math**

If the westbound car has gone a distance w, then w^2 + 24^2 = (3w+4)^2

**Math**

(1/2)(1/4)w = 1/4 ...

**Algebra 2**

Let the amounts be x,y,z. Then the facts are: x+y+z = 15000 y = x+2000 .05x + .06y + .04z = 730 Now just solve for x,y,z

**math**

depends on the base radius.

**Trigonometry**

no!! it is an ellipse. when t=0 you are at the right side, ready to move up and to the left: counterclockwise. http://www.wolframalpha.com/input/?i=plot+x%3D2cost%2B2,+y%3D3sint%2B3,+t%3D0+to+pi%2F2 better review how to plot points!

**Trigonometry**

well, x starts at a max and decreases y starts at the center point and increases. if that doesn't help, note that the ball is moving from (4,3) toward (2,6) care to try again?

**Math**

P(match) = 1/6 so, E = 20/6 * 5 - 20 ? -3 surprise, surprise -- the house wins. You only win 5/6 ticket per game. But it costs a whole ticket to play.

**sir steve help me reiny reiny**

small books! 10.98kg * 1book/54.9g = 10980/54.9 g/book = 200 g/book

**maths**

1 - 1/3 - (2/3)(2/3) = 2/9

**Math Trigonometry**

17sec^2(?)-13tan(?)sec(?)-15=0 17+17tan^2? - 13sec?tan? - 15 = 0 13 sec?tan? = 17tan^2? + 2 169 sec^2? tan^2? = 289tan^4? + 68tan^2? + 4 169tan^4? + 169tan^2? = 289tan^4? + 68tan^2? + 4 120tan^4? - 101tan^2? + 4 = 0 (24tan^2?-1)(5tan^2?-4) = 0 tan^2? = 1/24 or 4/5 ...

**math**

If it is now N, then in t years it will be N(1.02)^t

**Math**

(3/2 * 4/5)^2 = 36/25 = 1.44 so, 44% bigger

**math**

since AB=5, AC=12, BC=13 ABC is a right triangle. Similarly, ACD is a right triangle. So, the area should be easy to figure out, eh?

**math**

(a) 3 (b) 16 (c) 76

**math**

assuming you meant h(t) = 50+30t-5t^2 = 5(10+6t-t^2) max height is at t = -b/2a = 6/2 = 3 so, what is h(3)?

**physics**

pressure = force/area = N/m^2 so, plugging in your numbers, (1200*9.8)/(.06/4) = 784kPa or 115 psi If you mean each wheel is .06m^2, then divide that pressure by 4.

**maths**

3,4cm:17km = 1cm:5km = 1:500000

**physics**

r is distance (m), so assuming that x and y are also distance, a(m) = m, so a is just a constant b(s)(m) = m, so b = 1/s velocity = r' = ax' + by + bty' acceleration = r" = ax" + by' + by' + bty" = ax" + 2by' + bty" where ' ...

**Math**

(78/12 + 8/3) ------------------- = 55/23 (78/12 - 8/3)

**geometry**

try this: https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_23

**math**

(999-99)/11 = 81.81 so, there must be 81 multiples of 11 between 100 and 999 99+81*11 = 990 99+82*11 = 1001

**math**

http://www.wolframalpha.com/input/?i=plot+y%3Dx%2F(x%2B1)%5E2,+y%3Dx%5E5%EF%BC%8Dx

**Physics**

I'm having a bit of trouble with F. Force is in units of N = kg-m/s^2 Having v^2 there makes the units m^2/s^2. How do you get rid of the extra meters? Anyway, when you fix that, just remember that F = ma, so a = F/m. Just add that to the -9.8 m/s^2 of gravity, and find ...

**Maths**

the bug gains 1m each 2 minutes after 20*2 minutes it has climbed 20 meters During the 41st minute it climbs to 30m and is done.

**maths**

if there are b boys and g girls, b-1 = 2g 5(g-1) = b 5g-5 = 2g+1 3g = 6 g = 2 so, b=5 There are b+g=7 children

**math**

s:g = 3:2 = 9:6 g:r = 3:5 = 6:10 s:g:r = 9:6:10 9+6+10 = 25 so, we have s:g:r = 90:60:100 90+60+100 = 250

**math**

3b+4a+6m = 29 2b+6a+3m = 23 multiply the 2nd equation by 2, then add the two equations.

**math**

hits: 3*2 ? 5 7*6 ? 6 7*6 ? 4*7 2*5 = 5*2

**math**

3*60 = 2*90

**algebra**

How can 3 be a 2-digit number??? t = tens digit u = units digit so, the value of the number is 10t+u t+u = 6 10u+t = 10t+u + 36 The number is 15

**algerbra**

11s+4c = 93 9s+2c = 71 double the 2nd equation: 11s+4c = 93 18s+4c = 142 subtract the top from the bottom: 7s = 49 s = 7 11s+4c=93, so 77+4c = 93 4c = 16 c = 4 So, they sold 7 senior and 4 child tickets

**algebra**

11s+4c = 93 9s+2c = 71

**Math**

time = distance/speed so, figure average speed as (total distance)/(total time)

**Math**

correct. assuming that the ball isn't spinning in reverse, in which case it's green.

**Math**

not really. Aside from being incomplete, you did find that you can get 10 tickets along each side of the square. So, there are 10^2 = 100 tickets

**Math**

He can land anywhere he wants between A and B. If you want to find the path of least time, however, you need to minimize ?(x^2+12^2)/4 + (16-x)/8 where he lands x km from point A.

**Math**

why worry about the area? You are interested in the perimeter. One lap around the field covers 4*50 = 200 meters. Now, how many laps in 3000 meters?

**math algebra**

x+y = 400 .80x + .85y = 400-68 ...

**math**

why are you dividing? Does it even make sense to divide one dimension by another? The round cake has volume and area ?r^2h = ?(7/2)^2(2) = 76.97 in^3 ?r^2 + 2?rh = ?(7/2)^2 + 2?(7/2)(2) = 82.47 in^2 The rectangular cake has volume and area lwh = 6*9*2 = 108 in^3 lw + 2(lh+wh...

**math**

x/3 + 10 = 3x/4 4x + 120 = 9x 5x = 120 x = 24 The tank is 3/4 full, so 6 more gallons are needed.

**Calculus**

tossing in that extra x factor makes it beyond elementary functions, as shown here: http://www.wolframalpha.com/input/?i=%E2%88%AB(x+cos%5E2(x)sin(x))%2F(a-cos(x))+dx

**Calculus**

well, the first thing is to get rid of all those "nx" things. Let u = nx and we have 1/n ?(cos^2(u)sin(u))/(a-cos(u)) du now let v = a-cos(u) dv = sin(u) du cos(u) = a-v 1/n ?(a-v)^2/v dv Now it's simple.

**Math**

4000-460

**math**

the formula for y (distance) is bogus m/s^3 * t^2 = m/s the units don't work out right. It could be that you meant x = 4t + t^2 y = 2t + 2t^2 If so, then (a) set x=y and solve for t (b) set the speeds equal and solve for t

**math**

t:g = 13:8 g:b = 2:3 = 8:12 t:g:b = 13:8:12 ...

**calculus**

a good place to start is this site, with several useful examples. See what you make of it. http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

**Math**

well, duh: 2222 or 4444

**math**

If the quantities of coins are x = $1 y = $2 z = $5 then we are told x = 9z y = 8z x+2y+5z = 90 now solve for y

**English**

1 -- abandon, forsake

**algebra**

find the midpoint of the given segment: (0,-3) find the slope of the segment: -5/3 so, the slope of the bisector is 3/5 Now we have a point an a slope, so the perp. bisector is y+3 = 3/5 (x-0) or, in slope-intercept form, y = 3/5 x - 3

**Math**

clearly, the circle is (x+6)^2 + (y-7)^2 = r^2 So, I suspect a typo. Check your coordinates again, and recall that a circle with center at (h,k) is (x-h)^2 + (y-k)^2 = r^2 Once you fix your error, just find how far away it is from (4,-2).

**Math**

y+5 >= 7 subtract 5 from both sides: y >= 2 so, y could be less than 12 or greater than 12 for example, y=3 y+5 = 10 >= 7 y=20 y+5 = 25 >= 7

**math**

I assume you meant that each coin is 1/8 inch thick. Recall that the surface area of a cylinder is a = 2?r^2 + 2?rh = 2?r(r+h) Now just plug in your numbers.

**Math**

$k = 100k cents. So, p lbs costs 100k cents. Now you can easily see what 1 lb costs, right?

**Math**

well, he started out with 100d cents, right?

**math**

you might start here. Make a flowchart that reflects the logic of the code. http://www.jiskha.com/display.cgi?id=1495379575

**Math**

Let's start with Physics and Chemistry. at most 30% passed physics at most 28% passed chemistry so, at most 58% passed either of the two subjects. That is, the % who failed both is at least 42% That is 100-(30+28)% Now extend that to the other subjects.

**Math**

x/8 + 14 = x

**Math**

clearly a negative answer makes no sense. How did you arrive at your result?

**Math**

just like the goldfish problem. What do you think?

**Math**

suppose you divide a pound of tea in halves. How many are there?

**Math**

ok. how many half-pound pkgs in a whole pound? Multiply that by 2 1/2

**maths**

6.25^(1/2) = 2.5 2.5^3 = 15.625 or, using fractions, 6.25 = 25/5 (25/4)^(3/2) = 125/8

**Math**

Surely you can see that there is something missing where you say one this of the remainder earn salaries between 15000 and 18000 how much is "one this" ?? Do you not read what you post?