Saturday

November 22, 2014

November 22, 2014

Total # Posts: 26,888

**trigo**

First, vessels travel on headings, not bearings. 200 @ 245° = (-181.262,-84.524) 50 @ 320° = (-32.134,38.302) add them up to get (-213.396,-46.222) = 218 @ 258°
*September 16, 2014*

**Algebra**

correct.
*September 16, 2014*

**Calculus**

f(a) = 3a^2-2a f(a+h) = 3(a+h)^2 - 2(a+h) = 3a^2 + 6ah + 3h^2 - 2a - 2h f(a+h)-f(a) = 6ah + 3h^2 + 2h So, (f(a+h)-f(a))/h = 6a + 3h + 2 So, with various values of h, f(a+1) = 6a+3+2 f(a+0.1) = 6a + 0.3 + 2 = 6a + 2.03 and so on. Nothing difficult, just algebra I.
*September 16, 2014*

**Math**

That's a lot of work. Just think about it a bit. You start with 1 gram. 3 days later the amount left is 0.1 gram If the half life is k days, then (1/2)^3k = 1/10 3k = log(.1)/log(.5) = 3.32 So, k = 1.11 R(t) = (1/2)^(1.11t) But, since we like things base e for our ...
*September 16, 2014*

**Math**

just solve for x in the equation −0.01x2 + 130x − 180,000 = 174900
*September 16, 2014*

**math**

The expression factors as (3x-1)(x+2) However, there are lots of other ways to divide things up. For example, if x=3, the area is 8*5 = 40, but those are not the only two ways to get two numbers which multiply to 40.
*September 16, 2014*

**math help**

Just plug in the value: (4x+8)(x+2)(2x+4) / 2((4x+8)(x+2)+(4x+8)(2x+4)+(x+2)(2x+4)) = 8(x+2)^3 / 28(x+2)^2 = 2(x+2)/7
*September 16, 2014*

**Math**

steps to what?
*September 16, 2014*

**math**

The answer is r^2-5 + 5/(r^2-2) See the details when you enter your polynomials at calc101.com. Click on the long division button.
*September 16, 2014*

**Math**

after n hours, the length is 3*3^(n-1) = 3^n Now just plug in n=3
*September 16, 2014*

**math**

since the circumference is 8*3.14, the diameter is 8
*September 16, 2014*

**math**

f = m+5 f+5 = 2m m+5 + 5 = 2m m = 5
*September 16, 2014*

**math**

the first car went 20 miles before the 2nd car started out. So, it made up 20mi/(5/3 hr) = 12 mi/hr That is, car 2 was going 12 mi/hr faster than car 1, or 72 mi/hr Check: car 1 went 120 miles in 2 hours car 2 went 72(5/3) = 120 miles in 1 hr 40 min
*September 16, 2014*

**Algebra**

g(f(x)) = {g(8),g(5),g(-5),g(3)} = {10,9,3} g(3) is undefined
*September 16, 2014*

**Algebra 2**

y = 2x-1 + 3/5
*September 16, 2014*

**Algebra 1**

your solution is correct. To check on (180,-30), plug it in 8(180)+5(-30) =?= 900 But, how do you sell -30 student tickets?
*September 16, 2014*

**Math**

(a) max height occurs at t = -87/-32 (b) just solve for t when h=0
*September 16, 2014*

**Trig**

if the wind speed is w and the plane's speed is s, then (w cos-135°,w sin-135°)+(s cos63°, s sin63°) = (550 cos64°,550 sin64°) -.707w + .454s = 241.104 -.707w + .891s = 494.337 w = 31.089
*September 16, 2014*

**c**

or, 1/4 + 1/(x+9) = 1/x x = 3 so, the 2nd hose takes 12 hours check: in 1 hour, the two hoses fill 1/4 + 1/12 = 1/3 of the pool.
*September 16, 2014*

**college algebra**

If you mean the increase occurs only for those in excess of 3000, then you have (20)(3000) + (20+0.01)(x-3000) = 153525
*September 16, 2014*

**Coordinate Algebra**

In point-slope form, we have y-4 = -3(x-5)
*September 16, 2014*

**Math**

what, you don't know how to do any of those steps?
*September 16, 2014*

**Calculus**

Well, let's see what they have said. dL = k dT L0 so, L = k*L0*T + C Now you can plug in initial conditions or boundary conditions to determine k and C
*September 16, 2014*

**Math Proof:**

Looks to me like you just have to write the definition of inf and sup. a is not in E, but is greater than any other number less than any element of E.
*September 16, 2014*

**physics**

Looks like 600N/m^2 assuming it is lying flat on one face
*September 16, 2014*

**math**

1 = 1^3 8 = 2^3 27 = 3^3 so, what do you think?
*September 16, 2014*

**Math quadratic word problem**

Your very first line is off. For a triangle, A = bh/2
*September 16, 2014*

**Trigo**

Draw a diagram. If the pole has height h, and the tip of the shadow is x horizontally away from the pole, then (82 sin8° + h)/x = tan28° x/82 = cos8° Now, eliminate x to get (82 sin8° + h)cot28° = 82 cos8° Now just solve for h
*September 16, 2014*

**math**

2/5 + 1/3 = 6/15 + 5/15 = 11/15 That leaves 4/15 as woodwinds. So, 2/15 are clarinets 2/5 * 240 * 1/8 = 12 are tubas
*September 16, 2014*

**math**

0.9 x 5.2 / 0.2 x 25 = 585 (0.9 x 5.2) / (0.2 x 25) = 0.936 Pick the one you mean and then subtract 120 to find the missing value.
*September 16, 2014*

**physics**

distance = speed * time so, plug and chug
*September 16, 2014*

**math**

the only purpose I can see is readability. The math is identical, regardless of representation.
*September 16, 2014*

**Calculus**

It is not always possible to find such a tangent line. Consider the graph of y=x^2. No tangent line will pass through the point (0,1), since it lies inside the parabola. However, assuming that the point is properly placed, we need to find a point (x,y) on the graph such that ...
*September 16, 2014*

**Math Problem (please help)**

wolframalpha is your friend: http://www.wolframalpha.com/input/?i=6x^4+%E2%88%92+13x^3+%E2%88%92+144x^2+%2B+325x+%E2%88%92+150
*September 15, 2014*

**Calculus-Math**

Rolle's Theorem says that since C(2)=C(15), there is some value 2<c<15 such that C'(c) = 0. dC/dx = 12(2x^2-6x-9)/(x(x+3))^2 dC/dx=0 when 2x^2-6x-9=0, or x = 3/2 (1±√3) or, x = -1.1, 4.1 So, your answer of 4 is correct. The graph at http://www....
*September 15, 2014*

**math**

since 1/7 does not evenly divide 9/10, the answer is no.
*September 15, 2014*

**math**

#1 -14.8 + 22.6 = 7.8 #2 (x - 16) + 5 = -15 x-16+5 = -15 x-11 = -15 x = -4 #3 5 = –(–z + 3) 5 = z-3 8 = z #4 t - 17 = -11 Your choice is the situation where the temperature started at 17° and then decreased by t°.
*September 15, 2014*

**Math**

The number is negative, so it is -2.78 Its absolute value is +2.78 So, the distance between the two is 5.56 They are equal distances from 0, but in opposite directions.
*September 15, 2014*

**pre-calculus**

If there are x lamps, then 54x = 3564 So, there are 66 lamps sold If they cost c each, then 66c + 396 = 3564 So, the cost is 48. So, for 145 lamps, the profit is 145*(54-48) - 396 = 474
*September 15, 2014*

**Grade 11 Math functions**

All polynomials have domain xeR. All reals. Your domain makes no sense. xeR includes all values, including 2<x<3. Since y = (x - 5/2)^2 - 1/4, its minimum is at y = -1/4. So, the range is all reals y >= -1/4.
*September 15, 2014*

**geometry**

since the median is the middle value, the probability is 1/2 that a random value is above or below the median.
*September 15, 2014*

**math**

how long does it take to land? 9.8 - 4.9t^2 = 0 Now, with that t, how far does he go at 15.2 m/s? That would be 15.2 t
*September 15, 2014*

**algibra1**

If the 2nd side is x, then 2x + x + x+4 = 24 Now just find x, which is the 2nd side. Then it is easy to figure the other two sides.
*September 15, 2014*

**Pre-Calculus/Trigonometry**

correct
*September 15, 2014*

**Math**

use the law of cosines: d^2 = 18^2 + 11^2 - 2(18)(11)cos 70°
*September 15, 2014*

**Math**

actually, 711,311 is the only answer. If the hundreds digit is 3, then the remaining 4 digits must add to 4. They must all be 1's.
*September 15, 2014*

**dav**

the area scales as the square of the sides, so the sides are in the ratio √15/√25
*September 15, 2014*

**Math**

B clearly not, since 1+1=2, which is not in B T clearly, since adding any two multiples of 4 yields another multiple of 4. F yes, since adding any two integers yields another integer greater than 4. No idea. What is W?
*September 15, 2014*

**algebra**

If x is invested at 9%, then .068(4600) + .09(x) = .08(4600+x)
*September 15, 2014*

**algebra**

If x is drained, then .25(6-x) + 1.00(x) = .33(6)
*September 15, 2014*

**algebra**

if we call x the flying time for the 2nd plane, then since distance = speed * time, 660x + 540(x+1) = 1800 x = 1.05 hours So, how far did the first plane travel in 2.05 hours?
*September 15, 2014*

**Math 221**

(a) looks to me like it will be (30/110)(29/109)(28/108) Do (b) similarly. This is just like the problems involving drawing from a deck of cards.
*September 15, 2014*

**Math**

do the graphs at wolframalpha.com match your own? There, the two lines are clearly reflections in the x-axis.
*September 15, 2014*

**Math**

to reflect over the x-axis, y values change sign. So, f(x) -> -f(x), so it is -f(x) = 3x-5 Note that -f(-x) = -(-3(-x)+5) = -(3x+5) = -3x-5 See the reflection at http://www.wolframalpha.com/input/?i=plot+y%3D-3x%2B5%2C+y%3D3x-5
*September 15, 2014*

**algebra**

No sweat. Getting from the words to the algebra is half the battle. Do lots of word problems. They help you think mathematically.
*September 15, 2014*

**algebra**

If the cost is x, then x/4 = x/3 - 4000
*September 15, 2014*

**algebra**

I'll do one; you try the others. If you get stuck, come on back and show where. 5/13 t = -9 multiply by 13/5 to cancel the 5/13: t = -9(13/5) t = -117/5 I suspect #3 has a typo
*September 15, 2014*

**Algebra**

I think the easiest way is to use the point-slope form of the equation, since they gave you a point and a slope. Recall that the equation for a line through (h,k) with slope m is y-k = m(x-h) So, you have y+3 = -2(x+1) Now work with that: y+3 = -2x-2 y = -2x-5
*September 15, 2014*

**calculus**

(a) the period of sin(kt) is 2π/k, so we have period 2π/π = 2 seconds. (b) since the minimum of sin(kt) = -1, our minimum is -3/5 (c) since the length of the rod is 1, x^2+y^2 = 1 That means that y^2 = 1/25 (16 + 9cos^2(πt)) when x = 3/10, t=1/6, y = √...
*September 15, 2014*

**Maths**

since most bridges are level, I find the term questionable. I expect that in such problems, we're talking about the apex of the supporting structure, which is often a parabola. That would then be the vertex of the parabola.
*September 15, 2014*

**pre calc**

If x oz. of gold, we have 6000x + 50(1-x) = 200
*September 15, 2014*

**Pre Calc**

If normally the speed is x, then 90/x = 1 + 90/(x+3)
*September 15, 2014*

**Pre Calculus**

If the speed is x, then since time = distance/speed, 48/(x+4) = 32/(x-4)
*September 15, 2014*

**math**

Hint: Draw a diagram. You should recall that a common right triangle with integer sides is 7-24-25.
*September 14, 2014*

**AP Calculus**

You have y' = (2x-y)/(x-2y) at x = -2, y = -1±√13, so there will be two lines tangent to the ellipse at x = -2. for y = -1+√13, y' = (-4+1-√13)/(-2-2(-1+√13)) = (-3-√13)/(-2√13) = (13+3√13)/26 So, the line is y = (13+3&#...
*September 14, 2014*

**Algebra**

c+7 <= 3 c <= -4 So, plot a dot at -4 and shade everything to the left.
*September 14, 2014*

**Math**

since there are 10 choices for each digit, that would mean there are 10^7 possible numbers. Actually, SS numbers have 9 digits.
*September 14, 2014*

**Math**

Since there are 2 choices for each answer, there are 2^6 = 64 ways to answer all the questions. If some may be left unanswered, then we have 2^6 * 6C6 ways to select and answer all 6 2^5 * 6C5 ways to select and answer 5 2^4 * 6C4 ways to select and answer 4 and so on. Adding ...
*September 14, 2014*

**algebra**

just plug in either of the points. They must satisfy the equation, so if we pick (-8,-10), (-8+1.5)^2 + (-10+7)^2 = r^2
*September 14, 2014*

**algebra**

We need to rearrange terms into the standard equation: 5x^2−6x+5y^2+8y−9=0 5(x^2 - 6/5 x) + 5(y^2 + 8/5 y) = 9 (x^2 - 6/5 x + 9/25) + (y^2 + 8/5 y + 16/25) = 9/5 + 9/25 + 16/25 (x - 3/5)^2 + (y + 4/5)^2 = 70/25 Now you can read off the desired data.
*September 14, 2014*

**algebra**

domain: x >= 0 range: y >= 16
*September 14, 2014*

**Math**

If the original width is x length is 2x With a=2, our new box has volume (a)(x-2a)(2x-2a) = 32 2(x-4)(2x-4) = 32 2(x-4)(2x-4) - 32 = 0 4x(x-6) = 0 x = 6 So a sheet 6x12 will be cut to a box 2x2x8 with volume = 32
*September 14, 2014*

**Physics !!!!!!!!!!!!!!!!**

v(t) = Vo + at where a is negative, in this case. s(t) = Xo + Vot + 1/2 at^2 where a is negative If all you wanted was the formulas, why not just look in your text book? They are presented there. I'm sure.
*September 14, 2014*

**Limits**

1/(3+x) -> 1/3 1/(3x) -> ∞ clearly the sum is ∞ However, the left and right limits have opposite signs.
*September 14, 2014*

**PLEASE HELP! Math**

Liz is going 6 mi/hr faster than Mandy. By the time she starts, Mandy has gone (1/3)(30) = 10 miles So, at 6 mi/hr, it will take 10/6 = 1 hr 40 min to catch up with Liz. So, that will be 2 hours after Mandy started, or 1:00 pm. Or, you can plot both cyclists' positions as ...
*September 14, 2014*

**Math**

since each card has a 1/52 chance of being drawn, that would be (1/52)^12
*September 14, 2014*

**math**

75*42*(1/3) ft^3 * 7.5gal/ft^3 = 7875 gal
*September 14, 2014*

**Geography?**

Don't see much geometry in that ...
*September 14, 2014*

**Algebra 1**

start off with the point-slope form, since that is what they gave you: y+3 = -2(x+1) and rearrange things to fit the slope-intercept form.
*September 14, 2014*

**math**

well, just solve for x in -0.01x^2 + 130x - 180000 = 174900 x = 3900 or 9100
*September 14, 2014*

**AP CALC. AB**

#1: when is the denominator zero? When 1+sinx = 0 sinx = -1 x = 3π/2 + 2kπ Since f(-5) > 0 and f(-1) < 0, f(x)=0 only for a single value of x between -5 and -1. I assume the given choices included such a number.
*September 14, 2014*

**math**

2000000cm / (65π cm/rev) = 979.4 rev
*September 14, 2014*

**algebra**

Assuming you want to solve for x, x^2+x-1 = 0 looks like you need the quadratic formula. See http://www.wolframalpha.com/input/?i=x2+%3D+%E2%88%92x+%2B+1
*September 14, 2014*

**algebra**

(z-1)^2 = -4 z-1 = ±2i z = 1±2i
*September 14, 2014*

**algebra**

just solve -16t^2 + 200t = 380 -16t^2 + 200t - 380 = 0 Now just use the quadratic formula to find the roots.
*September 14, 2014*

**Pre-Calculus**

clearly, one such P(x) = (x-(2+i))(x-(2-i)) = ((x-2)-i)((x-2)+i) = (x-2)^2 - i^2 = x^2-4x+4 + 1 = x^2-4x+5 extra credit: why does the question imply that there is more than one such P(x)?
*September 14, 2014*

**physics**

it's probably easiest to find out how long it takes, then recall that v = at. So, if we have the car start the race at t=0, x=0, 35t + 3/2 t^2 = 48+24t t = 3.07 so, v = 3*3.07 = 9.21 m/s
*September 14, 2014*

**Math**

If the three E's are together, you might as well just count them as a single letter. So, there are just 6! ways to arrange the letters. Your 5!*6 is the same value. I assume you considered the 5! permutations of BRKLY times the 6 places you could stick the E's.
*September 14, 2014*

**Dang!**

Forgot the division step. Sorry.
*September 14, 2014*

**8th grade homework**

the value after each step is shown below. x x-6 10+x-6=30 so, x+4=30 x=26
*September 14, 2014*

**physic**

Just plug in your values and solve for y=0 in the equation of motion: y = 15 + x tanθ - g/(2(v cosθ)^2) x^2
*September 14, 2014*

**maths**

(1+5r) = 1.6 5r = .6 r = .12 or 12%
*September 14, 2014*

**Pre-Calculus**

start with point-slope form: y+3 = -2(x+1) Now just rearrange terms to the form you want.
*September 14, 2014*

**Algebra**

since time = distance/speed d/18 + d/10 = 14 d = 90 5 hours out, 9 hours back = 14 total
*September 14, 2014*

**math**

since cos 60.0˚ = 1/2, it has gone 24 km. 24km/1min = 24 km/min
*September 14, 2014*

**Math**

(2x-9)log7 = (2x)log4 2x log7 - 2x log4 = 9log7 x = 9log7/(2log7-2log4) There are, of course other ways of expressing those log terms...
*September 14, 2014*

**trig - eh?**

"3acos^2x.sin" means what?
*September 14, 2014*

**trig - eh?**

what does "a" have to do with anything?
*September 14, 2014*

**Maths**

80000/20000 = 4 so you want the time to grow by a factor of 4: (1+.10/12)^(12t) = 4 Now just solve for t.
*September 14, 2014*

**Pre-Calculus**

clearly you have x(x^2-x+1) Only one real root, of multiplicity 1. Two complex roots, also of multiplicity 1.
*September 14, 2014*

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