# Posts by steve

Total # Posts: 51,737

**Math**

just as with the others, the sum is a1/(1-r) just plug in your numbers.

**Math A Calculus**

this is calculus. Surely you can work with fractions... (a) a+b = (2/3)+(-1/4) = 8/12-3/12 = 5/12 (b) what do you think? recall that the length of ai+bj+ck = ?(a^2+b^2+c^2)

**Math**

so, what are the even multiples of 13? 26 52 78 so, what do you think?

**Math**

convert each displacement to rectangular form, and then just add them up. Then if the final position is (x,y), the displacement is ?(x^2+y^2)

**Math**

no diagrams here.

**maths**

well, 100% is all of the students. what is 100-54? no idea what the "strength" of a class is. Size? If so, then 23 = (1-.54)x

**geometry**

so, using the formulas, what do you get?

**Math (Sorta dumb question)**

Your final answer is correct.

**Math**

a1=243 r = 2/3 an = 243(2/3)^(n-1) a5 = 48 s = a/(1-r) = 5/(5/4) = 4

**Math**

hard to say, with no figure to compare ...

**Math**

recall that sin(90-?) = cos? cos(90-?) = sin? now see what you have.

**Algebra 1b**

(d) not (a) if n is negative

**Math**

huh? You surely must see from Scott's lines of text that 792 = 12C5

**Math**

oops tan(?) = y/x

**Math**

well, let's try to take this step by step. Let's call the two lower vertices L and R. Suppose we have L = (h,-k) Then the line LC has slope mLC = k/(1-h) and its equation (using point A) is y = k/(1-h) (x-1) Since LRC is equilateral, its angles are all ?/3. So, the ...

**Math**

well, let's try to take this step by step. Let's call the two lower vertices L and R. Suppose we have L = (h,-k) Then the line LC has slope mLC = k/(1-h) and its equation (using point A) is y = k/(1-h) (x-1) Since LRC is equilateral, its angles are all ?/3. So, the ...

**Math**

well, let's try to take this step by step. Let's call the two lower vertices L and R. Suppose we have ?/3 L = (h,-k) Then the line LC has slope mLC = k/(1-h) y+k = k/(1-h) (x-h) Since LRC is equilateral, its angles are all ?/3. So, the slope of LR is mLR = tan(arctan(...

**Calculus**

clearly, the series converges, since |sin(4n/?)+3|/4^n < |1+3|/4^n which converges to 16/3 But I can't come up with the actual limit value.

**Math: Trigonometry**

not quite -- 4/?7 is csc? did you draw the triangle?

**Math**

just set up equations to fit the facts. For example, #1 m/k = 5/6 m = k-12 so, (k-12)/k = 5/6 #2 7/15 = f/75 d = 75-f what do you get for the others?

**Math**

(c) because multiplying or dividing by a negative value changes the direction of the inequality 6 > 3 6(-4) = -24 < -12 = 3(-4)

**Math**

You know that Pn = 5+7n so, replace n with 4, and you get P4 = 5+7*4 = 33

**Math**

actual increase: 9.68-9.57 = 0.11 % increase: 0.11/9.57 = 0.01149 = 1.15%

**algebra**

15. yes 16. f(6) = 11*6+15 = 81 g(81) = -11*81^3 - 15*81^2 + 12*81 - 13 = -5,943,307 17. f(9) = 15*9^3 + 13*9^2 - 15*9 + 15 = 11,868 g(11868) = 15*11868 + 11 = 178,031 see? nothing to it! Although all the big numbers are annoying. The arithmetic just gets in the way of the ...

**algebra**

g(7) = -15*7^2 + 14*7 - 10 = -647 g(g(7)) = g(-647) = -15(-647)^2 + 14(-647) - 10 = -6,288,203 Not hard - just be careful with the arithmetic.

**help me maths anyone**

Draw a diagram. If we let T be the top of the mast O be the base of the mast P,Q be attachment points of two wires (PQ forms one side of the base of the triangle on the ground) M be the midpoint of PQ OV = 1 PV = 2 so, OP = ?3 triangle MOP is a 30-60-90 right triangle, so OP...

**@Dang**

Reiny is correct.

**sir steve damon reiny steve help maths**

Draw a diagram. Using similar triangles, if x = the distance from the lamp post s = length of shadow s/1.5 = (x+s)/6 s = x/3 so, ds/dt = 1/3 dx/dt ... Extra credit: how fast is the tip of the shadow moving?

**Algebra 2**

plus, since they are inverses, (b) is also correct.

**Algebra 2**

your answer is correct if you meant y = 2x y = log2x

**Algebra**

Who says it has to be simplified at all? However, since 97 is prime, 97x is a perfect square if x is 97. Or, if x is 97 times any perfect square.

**Math check my answer**

I agree... I think.

**Math~Algebra**

just plug in the numbers to the formula. Clearly you get (C).

**algebra help**

if the proper branch is taken, then (5,2) would be on the inverse. As it stands, the function has no inverse function, since it fails the horizontal-line test.

**Calculus**

at time t hours after the observation, the distance z between the ships is z^2 = (15-12t)^2 + (9t)^2 2z dz/dt = 90(5t-4) dz/dt=0 when t = 4/5 z(4/5) = 81

**Calculus**

(a) there is a local extreme every time sinx = cosx: x = ?/4 + k? (b) since the maxima occur when sinx=1, the envelope of these maxima is the curve y=e^-x.

**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