Monday

February 8, 2016
Total # Posts: 37,469

**Algebra**

they covered 450 miles in 6 hours. That's a combined speed of 75 mi/hr. So, if the slower train's speed is x, x + x+25 = 75
*November 30, 2015*

**math**

think of the size of a typical bedroom. Then think of the size of a typical sheet of paper.
*November 30, 2015*

**Math**

4x+40 = 10x+15
*November 30, 2015*

**Math**

or, solve by completing the square: x^2 - 34x = -100 34/2 = 17, so x^2 - 34x + 17^2 = 100 + 17^2 (x-17)^2 = 189 x-17 = ±√189 x = 17±√189
*November 30, 2015*

**Math**

Gotta use the quadratic formula for this one.
*November 30, 2015*

**geometry**

Call the vertex of the angle O Place the compass at O and draw a circle. Extend the sides of the angle so they intersect circle O at A and B. Place the compass at A and open it to the distance AB. Draw a circle with center A and radius AB. Circle A will intersect circle O at ...
*November 30, 2015*

**Geometry**

Mark off the data The triangles are both right triangles QS is a common side They have congruent hypotenuses Thus, they also have congruent sides PR and PS. (via the Pythagorean Theorem) So, by SSS, the triangles are congruent.
*November 30, 2015*

**Algebra 2**

seems simple enough to me. How many sets of 4 toppings are there? 10C4 = 210 Only one of those sets contains all 4 desired toppings. So, ...
*November 30, 2015*

**Math**

(5/6 + 5/6 + 5/6)/8 = (15/6)/8 = 15/48 = 5/16
*November 30, 2015*

**Maths:simultaneous equation**

clearly, x and y are non-negative. For, if x^3+y^3=1, then if x<0, y>1, and then x^4+y^4>1. Other than (0,1) and (1,0) I cannot think of any real solutions.
*November 30, 2015*

**History**

which amendment protects citizens' freedoms of speech and press a.1st b.3rd c.5th d.)7th its c?
*November 30, 2015*

**History**

Oh its A!
*November 30, 2015*

**Math**

Well, we know that √x has a domain of x>=0, so we could try √(x-3) but that has domain [3,∞) and we want to exclude 3. So, let's divide, since we cannot divide by zero: y = -1/√(x-3) Now 3 is excluded, and luckily, the range is now also correct. ...
*November 30, 2015*

**Geometry**

just draw a line parallel to one side, but almost on top of it. It is surely not the midsegment. In fact, of all the possible parallel lines, only one is the midsegment!
*November 30, 2015*

**Math**

if each first element is unique, it is a function.
*November 30, 2015*

**Math**

if it is a function, no two pairs have the same first element. So, ...
*November 30, 2015*

**Math**

Huh? You want to subtract 1/6 each time to get to the next term? I think you have the idea of common difference backwards.
*November 30, 2015*

**calculas**

you ant two rectangles with one side in common. So, if the big plot has dimensions x and y, then xy = 15000 Now, there will be two sides of length x and 3 sides of length y/3 (because of the extra fence down the middle), so the amount of fence is f = 2x+3y = 2x + 45000/x Now ...
*November 30, 2015*

**Math**

Ummm. how about C=0.20m+150.00 as they said.
*November 30, 2015*

**PreCalculas**

what, you can't compare your work to Damon's and see the differences? Hmmm.
*November 30, 2015*

**maths**

If d > 0, the sum is increasing if d < 0, the sum is decreasing So, if S20 = S30, d=0 That means that 20a=30a, so a=0 If a=0 and d=0, all the sums are zero.
*November 30, 2015*

**Math**

P(x) = R(x)-C(x) = x(100-10x) - (5x^2-800x+500) That' just a parabola, so its vertex is easy to find, no?
*November 30, 2015*

**Math**

since the slopes must be negative reciprocals (their product is -1), (3+5)/(3+4) * (-5-4)/(x+4) = -1
*November 30, 2015*

**math**

we may consider the height to start out at y=1. So, we want to know when 1-x/5 = 2.4 * (1-x/4)
*November 30, 2015*

**Algebra**

slope is change in y divided by the change in x. So, #1: (4-5)/(1-(-2)) #2: gotta do the division till you find the right pair. #3: change in y is 1/4 the change in x. So, since x changes by 4, y changes by 1... #4: since x has not changed, the line is vertical #5: again, ...
*November 29, 2015*

**Algebra**

check the solution in the related problems below.
*November 29, 2015*

**Pre-Calculus**

(a) surely you know where cos(x) = 0 (b) surely you can determine cos(0) (c) check the graph below (d) surely you know the min and max of cos(x) graph at http://www.wolframalpha.com/input/?i=cos%28x%29
*November 29, 2015*

**Intermediate Algebra**

(4,2)
*November 29, 2015*

**Calculus**

See whether you can get wolframalpha's result: http://www.wolframalpha.com/input/?i=%E2%88%AB+%28%28x-r%29^2+%281%2Fr%29+%28e^%28-x%2Fr%29%29%29+dx The limit would be undefined for x -> -∞ I suspect a typo somewhere. Seems like there should be e^(-x^2/r) or ...
*November 28, 2015*

**math**

good -- now you can work on your spelling...
*November 28, 2015*

**Math**

L{e^at f(t)} = F(s-a) so, L{e^4t f(t)} = F(s-4) now, what is L^-1(1/s^3) ?
*November 28, 2015*

**Math**

cos = 1/sec
*November 28, 2015*

**Algebra**

17n+11 - (5n+6) - (4n+5)
*November 28, 2015*

**Algebra 2**

b+m+p+n=17 200b+225m+175p+150n=3400 m = b+p+n-1 p = 2n Now just solve for the 4 unknowns.
*November 28, 2015*

**maths**

6200*.09*2.75
*November 28, 2015*

**Solving Trig Equations**

as you say, sinx = ±√3/2 You know the reference angle is π/3. sinx > 0 in QI, QII, so you have π/3 and 2π/3 sinx < 0 in QIII, QIV, so that gives you 4π/3 and 5π/3 But, for this problem, the domain is [-π,0] (QIII and QIV), so ...
*November 27, 2015*

**math**

scaling preserves angles, so all three triangles are similar.
*November 27, 2015*

**math**

S5 = 5/2 (2a+4d) S2 = 2/2 (2a+d) The sum of the last 3 is S5-S2, so we have S2 = 55-48 = 7 = 2a+d 2a+d = 7 5/2 (2a+4d) = 55 a=1, d=5 So, S3 = 3/2 (2+2*5) = 18
*November 27, 2015*

**Math**

the circumference is 8π ft. 2 ft per flower means you get 8π/2 = 4π ≈ 12 plants 20 seeds @ 2 ft/seed means a circle with a 40 ft circumference. So, the diameter is 40/π ≈ 12.75 ft
*November 27, 2015*

**Mathematics**

what is the formula for the ith term again? It seems garbled. There's a (3/n * floating around.
*November 27, 2015*

**Math**

y-intercept is where x=0. avg rate is (56-52)/(3-1) Looks like that would be 2 more hours. So, {0..5}
*November 27, 2015*

**Math (prob)**

the experimental results are his pass/fail of the test.
*November 27, 2015*

**Calculus**

let u = (x-r)^2 du = 2(x-r) dx dv = (1/r) e^(-x/r) dx v = e^(-x/r) ∫ u dv = uv - ∫ v du = (x-r)^2 e^(-x/r) - 2∫(x-r) e^(-x/r) dx Now do it all over again. Notice that the power of (x-r) has been reduced by one.
*November 27, 2015*

**math**

rotation of any multiple of 360/8 = 45 degrees. Also, a reflection across any of its 8 lines of symmetry.
*November 27, 2015*

**Math**

T(x) = 45 - x/300 (300,1) is not a point on the line. x has to increase by 300 for T to drop by 1°. Thus, (300,44) is the point you are after.
*November 27, 2015*

**math**

If there are x large prints, then there are 2x small prints. To break even, he needs 20*2x + 45*x = 510 I think you can now easily answer all the questions. Most of them don't even depend on the number of prints sold.
*November 27, 2015*

**math**

(A): the circle has a circumference of 8π feet. If the seeds are at least 2 ft apart, then at most 8π/2 = 4π ≈ 12 seeds may be planted.
*November 27, 2015*

**trigonometry**

since X is the foot of the altitude, any triangle with X as an angle will be a right triangle.
*November 27, 2015*

**trigonometry**

3-sided
*November 27, 2015*

**Trigonometry URGENT!**

amplitude is half the distance between the extremes. In this case, (125-99)/2 = 13 The midline (where sin(x) = 0) is (125+99)/2 = 112 The period is twice the time from max to min, or 28*2 = 56 So, we're looking at something like y=13 sin(pi/28 t)+112 Now for the phase ...
*November 27, 2015*

**Tamae**

220 * .05 = ?
*November 27, 2015*

**Math calculus**

see related questions below
*November 27, 2015*

**Integral Calculus**

if you set u = arctan(x), you have u du I expect you can handle that...
*November 27, 2015*

**jabu**

100 km/hr = 27.78 m/s v = √(2as) 27.78 = √(100a) a = 7.716 m/s^2 That is the minimum acceleration which will do the job. There is no maximum value.
*November 27, 2015*

**Math**

If the markup is x%, then (1+x/100)*.84 = 1.12 x = 33.33% = 1/3 1200 * 4/3 = 1600 check: 1600*.84 = 1344 = 1200*1.12
*November 27, 2015*

**Precalculus ????**

23 is the maximum hours, but you want the maximum temperature. sin(x) has a maximum of 1 5sin(x) has a max of 5 5sin(x)+19 has a max of 24 It appears there is a typo, since 24 is not a choice.
*November 27, 2015*

**math**

You have 1-2 + 3-4 + ... + 99-100 = (-1) + (-1) + (-1) ... + (-1) (50 times) = -50 Naturally, the 2nd sum is then +50
*November 27, 2015*

**math**

one step at a time... 11/15=(1/a+(b/c+(d/e+f))) 11 = 15(1/a+(b/c+(d/e+f))) 11 = 15/a + 15(b/c+(d/e+f)) 15/a = 11 - 15(b/c+(d/e+f)) a = 15/(11 - 15(b/c+(d/e+f))) Now for b: 11 = 15/a + 15(b/c+(d/e+f)) 15(b/c+(d/e+f)) = 11 - 15/a (b/c+(d/e+f)) = 11/15 - 1/a b/c = 11/15 - 1/a - (...
*November 27, 2015*

**math**

If there are n survivors, there are 5n meals. After 2 days, 2n meals have been eaten, leaving 3n Then the 3n meals lasted the n-4 survivors 5 days: 3n/(n-4) = 5 Now just find n
*November 27, 2015*

**math**

list the powers of 2 and their remainders: 2^1 2 2^2 4 2^3 1 2^4 2 2^5 4 2^6 1 ... you can see a pattern. So see how many times the pattern is repeated in 1000 powers of 2, and how many powers are left over.
*November 27, 2015*

**Precalculus**

no ideas of your own to provide? These two problems are really the same thing with different numbers. As you might imagine, the trick is to come up with a function you can analyze. So, on the first one, if we are starting at the midline, then since sin(0) = 0, we expect to ...
*November 26, 2015*

**Math - PS**

skip that middle section. It was a blind alley I forgot to erase. Cool problem. I had to try several things before I hit on the right trick to convert all the sums to products.
*November 26, 2015*

**Math**

sin2x+cos2x = √2(sin2x * 1/√2 + cos2x * 1/√2) = √2sin(2x+π/4) so you have [√2sin(2x+π/4)+1]/[√2sin(2x+π/4)-1] = [√2sin(2x+π/4)+1]^2/(2sin^2(2x+π/4)-1) Now, using sum-to-product formulas, √2sinu+1 = √...
*November 26, 2015*

**maths.augusco**

1/5 : 3/10 : 7/1 = 2/10 : 3/10 : 70/10 = 2:3:70 There will be some pieces in the mix, as 75 does not divide 720 I suspect a typo. Fix it, and you will see a better division of apples.
*November 26, 2015*

**discriminant**

24
*November 26, 2015*

**c programming**

what is the difficulty? Do you have trouble with I/O? Don't understand arrays? Don't understand functions? Have you worked out pseudo-code that takes the necessary steps?
*November 26, 2015*

**Programming**

well, just follow the loop. On entry, count=1 next iteration: count steps up by 2, so count=3 next iteration: count=5, so exit so, the output is 1 3
*November 26, 2015*

**Pre Calculas**

In fact, ln56/ln7 = log756
*November 26, 2015*

**math**

true
*November 26, 2015*

**calculus**

The first step is just to expand the polynomial: e^2t + 2 + e^-2t Now integrate each term. Or, remember that your function is just (2 cosh t)^2 = 4cosh^2(t) now use the half-angle formula for hyperbolic functions, and you will get the same answer. Hmm. I guess just expanding ...
*November 25, 2015*

**math**

since time = distance/speed, 4/(x-2) = 12/(x+2) Now just find x.
*November 25, 2015*

**math**

see related questions below.
*November 25, 2015*

**maths**

1/3 * 3/4 (1 is not prime)
*November 25, 2015*

**Algebra**

the growth after t hours is 1.036^t so, find t where 1.036^t = 2 t log1.036 = log2 t = log2/log1.036 = 19.5986
*November 25, 2015*

**Geometry**

If the answer was (2,5), the line of reflection was y=4. That, or A=(-3,-3)
*November 25, 2015*

**Geometry**

(-3,3) -> (2,3) -> (2,-1) or, (-3,3) -> (-3,-1) -> (2,-1) typo?
*November 25, 2015*

**Algebra**

exclude values where the denominator is zero. So, where is (k+7) = 0?
*November 25, 2015*

**math**

-sin10°
*November 25, 2015*

**just checking(:**

Bzzt. But thanks for playing It is b) -2 -4y = 8x y = (8/-4)x = -2x
*November 25, 2015*

**Algebra 1 Honors**

2.50 + 0.35(15.75-1)*4 = $23.15
*November 25, 2015*

**math**

it occurs where d/dx(R-C) = 0 I get a maximum profit of 3992.81 at x=2875
*November 25, 2015*

**Trig**

This is not an identity. It is an equation to solve. cos 2x = -.3 cos 72.5° = 0.3 But, we have -.3, so 2x is in QII or QIII 2x = 107.5° or 255.5° x = 53.75° or 127.75° You can also add 180° to each of those, since cos(2x) has period of pi. That will ...
*November 25, 2015*

**math**

clearly, y=x
*November 25, 2015*

**science**

what is the speed of sound at 25°C? If that speed is s, then If the cliff has height, it hits the bottom at time t, where 4.9t^2 = h. So, h = (8-√(h/4.9))*s Plug in your speed s, and find h.
*November 25, 2015*

**math - eh?**

Better read your question. It is gibberish.
*November 25, 2015*

**math**

18 * (32/72) 30 * (8/12)
*November 25, 2015*

**Math**

true, false if x has powers, it is a polynomial. If x is the power, it is an exponential, as in e^x polynomials have constant powers of a variable. exponentials have variable powers of a constant.
*November 25, 2015*

**Pre Calculus**

for any positive real a, y = a^x always goes through (0,1), so y = 4a^x always goes through (0,4) y = a^x + 3 also goes through there. y = h*a^x + k too, if h+k=4 negative powers also work.
*November 25, 2015*

**Algebra**

If x hours at 57 mi/hr, then since distance = speed * time, 57x + 56(4.5-x) = 254
*November 25, 2015*

**Geometry**

yes.
*November 25, 2015*

**Calculus 1**

You can see a nice graph with grid lines here: http://rechneronline.de/function-graphs/ You may have to adjust the y scale some so it fits. You can use that to estimate the area. I'm sure you can then evaluate the integral.
*November 25, 2015*

**derivatives**

is that (cos t)^2 or cos(t^2) ? Then express t as functions of x and y, and you have dy/dx = (dt/dx) / (dt/dy) then you can use that to get the 2nd derivative
*November 25, 2015*

**electron theory and ohms law**

ummm. E = IR ? This time find I.
*November 25, 2015*

**electron theory and ohms law**

E = IR = 6.4 * 20
*November 25, 2015*

**math**

average cost is (1/3 x^3 – 18x^2 + 160x)/x = 1/3 x^2 - 18x + 160 marginal cost is x^2-26x+160 so, where is 1/3 x^2 - 18x = x^2 - 26x ?
*November 25, 2015*

**Algebra II, part 2**

#6. Surely by now you can do this... #7. b^2-4ac = 9+4*2*18 #8. 3√-3 = 3√3 i #9. Multiply top and bottom by conjugate: 8(3-2i)/13
*November 24, 2015*

**Algebra II**

#1. x^2 has negative coefficient. downward #2. (5/2)^2 #3. hint: 15+1 = 16 #4. does not factor. Did you mean +2x? #5. shift left 2, scale vertically by 1/3, shift up 5
*November 24, 2015*

**algebra 2**

Well, I see 2x^2+3x+1 and 3ay^2+9a If you want to factor them, then you have (2x+1)(x+1) and 3a(y^2+3) Other than that, I'm not sure what you want to do with them.
*November 24, 2015*

**Trigonometry**

#1. no calculator? It will be in 1st quadrant. #2. how long is the court? The angle is x where tan(x) = 2.44/(court length)
*November 24, 2015*

**Math**

so, if the width is w, 2(w + 2w) = 126
*November 24, 2015*