Friday

September 30, 2016
Total # Posts: 44,265

**maths**

Set up the proportions: 9/x = 75/10 x/2.5 = 75/10

*September 26, 2016*

**Engineering Math Growth**

The logistic function is y = a/(1+be^(-kx)) y(∞) = 300 ==> a = 300 y(0) = 40 ==> 300/(1+b) = 40 b = 15/2 The k just affects how slowly the population approaches the final value. See http://www.wolframalpha.com/input/?i=plot+y%3D300%2F(1%2B7.5e%5E-x),+y%3D300%2F(1%...

*September 26, 2016*

**Math**

yep But I'd just leave it as √85 That is its exact value, not some decimal approximation.

*September 26, 2016*

**Calculus**

google is your friend. You might want to start here: http://math.stackexchange.com/questions/1942079/constructing-a-function-with-saddle-points

*September 26, 2016*

**Translate to an algebraic expression**

q=x

*September 26, 2016*

**mathematics need help me**

google is your friend. By searching for "sum 1/(n+k)" I found http://math.stackexchange.com/questions/285308/limit-of-the-sum-frac1n1-frac1n2-cdots-frac12n

*September 26, 2016*

**calculus**

∫dx/((x+1)√(1-x^2)) I'd try a trig substitution on this one x = sinu dx = cosu du √(1-x^2) = cosu Now you have ∫(cosu du)/((1+sinu)cosu) = ∫1/(1+sinu) du = ∫(1-sinu)/(1-sin^2u) du = ∫(1-sinu)/cos^2u du = ∫sec^2u du - ∫...

*September 26, 2016*

**Pre-calc**

the domain of √u is u>=0 So, we need x+h >= 0 That means h >= -x and h≠0

*September 26, 2016*

**Physics**

Draw your triangle. It should be clear that tanθ = x/y watch the signs of x and y to get the proper quadrant for the angle.

*September 26, 2016*

**Riddles**

farmers

*September 26, 2016*

**Algebra**

20.00 + .10(26457/60) usually they charge for each additional minute of part thereof, so you need to round 26457/60 up to the next minute

*September 26, 2016*

**Physics**

This reduces to finding the max height of a ball thrown upward with a speed of 12.0 sin65° m/s I'm sure you've done that kind before.

*September 25, 2016*

**Math**

(4/5 -3/4) * 300

*September 25, 2016*

**Math**

only for add and subtract

*September 25, 2016*

**precalculus - incomplete**

not much to go on, one side.

*September 25, 2016*

**Chemistry**

the mole ratio is very nearly 1:1, so your answer seems good.

*September 25, 2016*

**Physics**

how far can you jump with that speed and angle? The range is R = v^2/g sin2θ Now that you know the distance, figure your time in the air, with a constant speed of v cosθ. Now, how far does the board go in that time? You can't be farther away than the difference ...

*September 25, 2016*

**Physics**

You know that the height h is h = v*sinθ t - g/2 t^2 So, solve for g in 21.0 sin26.0° * 7.3 - g/2 * 7.3^2 = 0 g = 2.52 m/s^2

*September 25, 2016*

**phsyics**

the magnitude is √(Fx^2 + Fy^2) So, figure the min and max values for F and use the average ± something

*September 25, 2016*

**Calculus**

Just check the limits at the interval endpoints. If the limits are the same, f(x) is continuous there.

*September 25, 2016*

**Math**

Take a look at bij as a sum of elements of A and AT

*September 25, 2016*

**Math**

ummm 10+8 ft^2/min

*September 25, 2016*

**polynomials**

First of all, it's not a polynomial,as there is only one term. It's ok to here: (9x^24y^18)(1/3xy) Now, is that 1/(3xy) or (1/3)xy ? It appears that you meant (1/3) xy, but it's hard to say, since you some how got (...)(1/3) = 3(...) How did you manage to switch ...

*September 25, 2016*

**Math**

assuming all the balloons have equal chance of getting hit, this is just the same as choosing without replacement: p = 4/15 * 3/14

*September 25, 2016*

**alg 2**

If their speeds are x and x+6, then since distance = speed * time, (x + x+6)(7/2) = 273

*September 25, 2016*

**Calculus**

First, I think you need to include x=0 in only one branch. f(0) = √(x+a) + 1 = 1+√a So, we need 2x^2+3 = 1+√a at x=0 That makes a=4 Now we need f(1) to match up: 2x^2+3 = (x+b)^2+1 2+3 = (1+b)^2+1 4 = (1+b)^2 b = 1 or -3

*September 25, 2016*

**physics steve or damon help!!!!**

well, 6:70 is not even a valid time, but assuming it's minutes and seconds, 6:70 = 490 seconds. Looks like 2 SD

*September 25, 2016*

**Math**

1/4 w + 3 < -6 1/4 w < -9 w < -36

*September 25, 2016*

**calculus**

review the conditions for the series to exist. In this case, it will exist for x=0, so we want y = ∑anx^n y' = ∑nanx^(n-1) y" = ∑n(n-1)anx^(n-2) The DE then says ∑n(n-1)anx^(n-2) + x*∑nanx^(n-1) + ∑anx^n = 0 ∑n(n-1)anx^(n-2...

*September 25, 2016*

**Physics**

x^2 + (2x)^2 = 112 The angle θ West of North is tanθ = 2/1

*September 25, 2016*

**math**

m = ∆y/∆x = (-6-(-4))/(7-4) = -2/3 I have no idea what you did wrong, since you did not show your work. The other thing you did wrong was misspell "wrong" -- twice!

*September 25, 2016*

**Maths**

4x^2 - 6y^2 = 24 8x - 12y y' = 0 y' = 2x/3y So, at any point (x,y) on the hyperbola, the slope is 2x/3y. The line 2x+c has slope 2. So, where do we have 2x/3y = 2? x = 3y 4x^2-6y^2=24 4*9y^2-6y^2=24 30y^2=24 y^2=4/5 y=2/√5 Things are lookin' good, with that...

*September 25, 2016*

**Math**

h = 3j h-3 = 4(j-3) now crank it out

*September 25, 2016*

**Math Algrebra Word Problems People on a line**

Draw a line, and label the people and their distances. #1 The 4000th person from the front is the last person. So, between that person and the 1000th from the last there are 999 people. Now do some thinking and try the others. Why do I think that it is not (just) the teacher ...

*September 25, 2016*

**Physics**

1 km/hr = 1000/3600 = 5/18 m/s time taken to decelerate: t = (142.5-36.7)(5/18)/1.6 = 18.368 s In that time, it traveled s = 142.5t - 0.8t^2 meters

*September 25, 2016*

**Expected Value**

(1/40)(4) + (10/40)(2) - (29/40)(1) = -5/40

*September 25, 2016*

**maths**

as you know the hyperbola y^2/a^2 - x^2/b^2 = 1 has asymptotes x = ±b/a y

*September 25, 2016*

**maths need help**

in standard form, your hyperbola is 3x^2-y^2+2y-1=0 3x^2 - (y-1)^2 = 0 x^2 - (y-1)^2/3 = 0 Your hyperbola is just two intersecting lines, so its transverse is zero! I suspect a typo.

*September 25, 2016*

**math**

f(x) =x(x+2) f(-x) = (-x)(-x+2) f(x)-f(-x) = x(x+2)- (-x)(-x+2) = x(x+2) + x(-x+2) = x(x+2-x+2) = x(4) = 4x

*September 25, 2016*

**Math - eh?**

You say f'(x) >= f(x) > 0 for all x in [0,∞) show that f'(x) >= f(x) for all x in [0,∞) Is that not the hypothesis already?

*September 25, 2016*

**maths need help**

f = 3r f+8 = 5/2 (r+8) r=24, so f=72 check: (72+8)/(24+8) = 80/32 = 5/2 So, (72+16)/(24+16) = 88/40 = 11/5

*September 25, 2016*

**English Grammar**

for, because, since and we usually say "The sun"

*September 25, 2016*

**Maths**

2 1/6 - 1 1/3 = 13/6 - 8/6 ...

*September 25, 2016*

**math**

195 is 65% of 300, not of 3000 Anyway, that's not the question. 65% are apples 25% are pears So, 40% difference, means .40*3000 = 1200 more apples than pears

*September 25, 2016*

**maths**

tanA + 4/tanA = 4 (tan^2A+4)/tanA = 4 tan^2A - 4tanA + 4 = 0 (tanA-2)^2 = 0 tanA = 2 ...

*September 25, 2016*

**Science**

(65-16)km/hr / 10min = 4.9 (km/hr)/min now, 1 km/hr = 1000m/3600s, so now you have (4.9 * 1000/3600 m/s) / 60s = 4.9*1000 / 60*3600 m/s^2

*September 25, 2016*

**Chemistry**

Similar question to http://www.jiskha.com/display.cgi?id=1474753951 "If that 22.33g sample of calcium chloride was dissolved in 1000.0g of water, with both substances originally at 25.0C, what would be the final temperature of the solution be? Assume no loss of heat to ...

*September 24, 2016*

**Physics**

The range is v^2/g sin2θ. So, solve for θ in 15000^2/16 sin2θ = 1200 * 5280 for the time, use θ to find when h=0: h(t) = 15000sinθ t - 8t^2 = 0

*September 24, 2016*

**Applications of functions**

y=20-x

*September 24, 2016*

**Precalculus**

No idea of your diagram, but this illustration is found in many places. Probably even in your own textbook. Surely online.

*September 24, 2016*

**Math/Chem**

yes

*September 24, 2016*

**Physics**

You need to know how far each moved during the jump. You know the speeds, so you just need to know how much time the jump took: v sinθ t - 4.9t^2 = 0 Solve for t and use it to get the distances.

*September 24, 2016*

**Physics**

google four point charges field at midpoint and you will find many similar problems. Just tweak the numbers.

*September 24, 2016*

**Physics**

you don't really need to write in the arrows. Anyone familiar with the topic knows these are vector quantities. Torque = p x E you have p, so how is E defined in terms of Q and r?

*September 24, 2016*

**science**

momentum = mass * velocity

*September 24, 2016*

**Maths**

Hmmm. I'd say 63 64 61

*September 24, 2016*

**Math**

mark your point: 8 On the number line, "less than" means "to the left of" So, in this case, that means toward the origin, and to infinity and beyond ...

*September 24, 2016*

**Spanish**

I'm sure you can, after a little research.

*September 24, 2016*

**Math**

0 <= 2 is true 0 is less than 2 0 is not equal to 2 zero IS less than OR equal to 2 If you say that today is hot or humid, it is true even if it is not humid, as long as it is hot.

*September 24, 2016*

**Physics**

How long were you in the air? v sinθ t - 4.9t^2 = 0 So, the dog's speed is 1.72/t This assumes that at t=0 you and the dog were in (approximately) the same spot.

*September 24, 2016*

**Chemistry**

with constant V, P/T is constant. So, what is P/T at stp? Now substitute in your new P.

*September 24, 2016*

**Math**

ok on both

*September 24, 2016*

**Physics**

the distance Chloe jumps is v^2/g sin2θ Now you know her horizontal speed: v sinθ how long she was in the air: v sinθ t - 4.9t^2 = 0 you can figure how far they both moved in that time.

*September 24, 2016*

**Physics**

Starting at the top of the ramp, the rider's height is h(t) = 1.60 sin23° t - 4.9t^2 So, find t when h=0 The rider's velocity when he hits the ground is Vx = 1.60 cos23° Vy = 1.60 sin23° - 9.8t So, the speed v at time t is found using v^2 = Vx^2 + Vy^2

*September 24, 2016*

**maths**

google is your friend: (a) https://answers.yahoo.com/question/index?qid=20131112055820AAxrUAf (b),(c) Check out Riccati equations y' = y^2 + q(x)y + p(x)

*September 24, 2016*

**Maths Percentage**

22/24 * 100 = ?

*September 24, 2016*

**Math**

If the length of AB is x, then AP = 2x/5 AQ = 3x/7 PQ = AQ-AP = 3x/7 - 2x/5 = 2 Now just solve for x

*September 24, 2016*

**maths**

infinitely many. Any point on the x-axis will do.

*September 24, 2016*

**math**

geez. How many ways are there to add 18 + 16?

*September 24, 2016*

**Physics**

340m/s * 2s = 680m

*September 24, 2016*

**Math**

48/(5+(11-8)) - 7

*September 24, 2016*

**Maths - Woe Is Me!**

fix the typos And what does p>+-24 mean. I get that +- means ±, but what does p>±24 mean? Also what are you supposed to do with 18x^2+2px+32 ?? factor it? solve for the roots? find p such that it is a perfect square? what, oh, what? Wooooeeee is meeee!!

*September 24, 2016*

**Geometry**

You know that AP+PQ+QB = a Now, AP = 2PQ = 2BQ That means that PQ = QB and that gives us 2PQ + PQ + PQ = a 4PQ = a PQ = a/4 Let M be the midpoint of QB Let N be the midpoint of AP AM = AP+PQ+ QB/2 = a/2 + a/4 + a/8 = 7a/8 NM = AP/2 + PQ + QB/2 = a/4 +a/4 + a/8 = 5a/8 Draw and ...

*September 24, 2016*

**Geometry**

first, recall that supplementary means they add to 180. So, now you know that A) The two angles are x and x-40 x-40 + x = 180 B) The two angles are 5x and 4x. So, 5x+4x = 180 So, just solve for x, and then you can work out the angles.

*September 24, 2016*

**Trigonometry**

see related questions below google can also help out

*September 24, 2016*

**maths~log**

just for ease of readability, let's say u = log5x v = log3y Then we have 2u+3v=8 6u+2v=2 This has the solution u = -5/7 v = 22/7 Unusual for a problem of this kind, but anyway... log5x = -5/7, so x = 5^(-5/7) log3y = 22/7, so y = 3^(22/7)

*September 24, 2016*

**maths**

I think your sum-to-product formulas will make this just drop out easily.

*September 24, 2016*

**Maths**

AB^2 + BC^2 = AC^2 1 + 1 = 2

*September 24, 2016*

**Maths**

assuming this is an isosceles triangle, I'd say they are both (180-65)/2

*September 24, 2016*

**math**

6 1/2 + 7 2/3 = 6 3/6 + 7 4/6 = ...

*September 24, 2016*

**Algebra**

plug in your data at http://www.gregthatcher.com/Mathematics/GaussJordan.aspx for the details, and confirmation of your answer >wink wink<

*September 24, 2016*

**Math**

DNE means Does Not Exist, so (a) is the choice of choice.

*September 23, 2016*

**Physics**

the range R = v^2/g sin2θ Now just plug in your values

*September 23, 2016*

**Chemistry**

25/800 = 1/32 = 1/2^5 so, it will take 5 half-lives

*September 23, 2016*

**Math**

Try plugging in you numbers here and see whether it is the same. The range is R = v^2/g sin2θ

*September 23, 2016*

**Math**

so, what do you get?

*September 23, 2016*

**Math**

√n is defined only for n>=0 So, the domain of √(2x+1) is (2x+1)>=0, or x >= -1/2 similarly for the other. There is always a domain.

*September 23, 2016*

**Math**

so, after t years, it is 41250 * 1.02^t

*September 23, 2016*

**Physics**

F = ma = 0.39 * 930√2

*September 23, 2016*

**Math , trigonometry**

h/20 = tan(47)

*September 23, 2016*

**physics**

F=ma plug in your numbers and evaluate a. Watch the units.

*September 23, 2016*

**Newtons Cooling Method**

-1.002 ≈ 1, so let's use T(t) = -53e^-t + 89.50 = 85.25 53e^-t = 0.25 e^-t = .004717 t = -ln .004717 = 5.3566

*September 23, 2016*

**math**

−2(x+2/3)+1=5 −2(x+2/3) = 4 x + 2/3 = -2 ...

*September 23, 2016*

**ODE Piecewise**

y'-y = 1: y = ce^x - 1 for 0<=x<=1 y'-y = -1: y = ce^x + 1 for x>1 y(0) = 1 means c=2, so y = 2e^x - 1 for 0<=x<=1 If we want y to be continuous, then since y(1) = 2e-1, y = (2 - 2/e)e^x + 1 for x>1 Thus, y(2) = (2 - 2/e)e^2 + 1 = 2e^2-2e+1

*September 23, 2016*

**Engineering Math**

Start with Newton's Law of cooling. Using your numbers, with To the initial temperature, T(t) = 10 + (To-10)e^(-kt) Now you have T(1) = 10 + (To-10)e^-k = 40 T(2) = 10 + (To-10)e^-2k = 25 (To-10)e^-k = 30 (To-10)e^-2k = 15 (To-10)^2 e^-2k = 900 (To-10) e^-2k = 25 Now ...

*September 23, 2016*

**5th grade math**

14+2=6+2x(3+2)

*September 23, 2016*

**Engineering Math, Newtons Cooling Method**

Your DE yields v = c e^(-2t) + 4.9 v(0) = 0 means c = -4.9, so v(t) = 4.9 - 4.9e^(-2t) (a) plug in t=1.5 (b) v(∞) = 4.9

*September 23, 2016*

**Physics**

v = 30*sin30° - 9.8t it reaches max height when v=0.

*September 23, 2016*

**maths**

Man. some messy algebra! Clearly there are horizontal asymptotes at y = ±b and curve passes through (a,0). The shape seems to preclude any simple equation or parametric equations. Maybe more later.

*September 23, 2016*