Friday

October 9, 2015
Total # Posts: 34,421

**maths**

How about 1 and 6? -6 + 1 = -5 That help?
*September 16, 2015*

**maths**

which two factors of 6 differ by 5?
*September 16, 2015*

**maths**

Expand and collect terms: (x-2)^2+3x-2=(x+3)^2 x^2-4x+4+3x-2 = x^2+6x+9 x^2-x^2-4x+3x-6x+4-2-9 = 0 ...
*September 16, 2015*

**Physics**

In x-y coordinates, A = (-13.156,4.788) B = (0,11) C = (-9.830,-6.883) D = (16.070,-19.151 Add them all up and you get (-6.916,-21.246) The resultant thus has magnitude √(6.916^2 + 21.246^2) = 22.343 and direction θ where tanθ = -21.246/-6.916 = 71.97° in ...
*September 16, 2015*

**science**

momentum before: mv momentum after: m(-v) now just subtract to get the change.
*September 16, 2015*

**maths**

which two factors of 18 add to 11?
*September 16, 2015*

**maths**

suppose you had 2a-ta Could you factor that? If so, just replace a with (x+y) in this problem.
*September 16, 2015*

**maths**

You didn't do any factoring. 3xsquared y-9xysquared+12xcubed y cubed = 3x^2y - 9xy^2 + 12x^3y^3 = 3xy(x - 3y + 4x^2y^2)
*September 16, 2015*

**maths**

x^2/2 + 2x^2/3 - 7x^2/6 x^2(1/2 + 2/3 - 7/6) ...
*September 16, 2015*

**maths**

(x^2-4x)/(x^2-2x-8) (x)(x-4) / (x+2)(x-4) ...
*September 16, 2015*

**Geometry**

Draw a diagram. Clearly m∠PQR = m∠PQS + m∠SQR
*September 16, 2015*

**Math**

well, 2*8 = 16 ...
*September 16, 2015*

**Math**

Is there a question in there? As it stands, the two equalities are inconsistent.
*September 16, 2015*

**physics**

clearly, the speed is √(34.1^2 + 53.5^2)
*September 16, 2015*

**physics**

1.3m/s * 148s * sin 27° = 87.3m
*September 16, 2015*

**Algebra**

twice a number: 2x increased by 28: ...
*September 16, 2015*

**math**

120000*(1.07)^1.5
*September 16, 2015*

**chemistry**

maybe
*September 16, 2015*

**trigonometry**

since s = rθ, you want r when 12(π/180)r = 59
*September 16, 2015*

**college algebra**

if the width is w, the length is 2w+2. So, you have 2(w + 2w+2) = 64 Now just find w and then the length.
*September 16, 2015*

**Chemistry**

(C)
*September 16, 2015*

**Math-pre algebra**

you are given so little data, the expression must be quite simple. How about 84:162
*September 16, 2015*

**Chemistry**

25*.1 + 50*.05 = (25+50)*x Where did you get that 5?
*September 16, 2015*

**Math**

well, 1/10 = 0.1 So, what do you think?
*September 16, 2015*

**Math**

3.024
*September 16, 2015*

**Math**

five hundred eight - one thousandth is written as 0.508 So, what is 10 times that?
*September 16, 2015*

**geometry**

The circle is (x-8)^2 + (y-2)^2 = r^2 Plug in your point (3,7) and you have (3-8)^2 + (7-2)^2 = r^2 Now that you have r^2, you know the area is just pi r^2. If you get stuck, come back with your work.
*September 15, 2015*

**Geometry**

find the lengths WX and XY. Twice their sum is the perimeter. Use the distance formula: WX = √((-5-(-3))^2+(4-7)^2) = √(2^2+3^2) = √13 and so on
*September 15, 2015*

**Math**

1 x 7 = 7
*September 15, 2015*

**Math**

Since each foot requires 4 tiles, the table (as measured in tiles, instead of feet, is) (5*4)(8*4) tile^2
*September 15, 2015*

**calculus**

well, (-2+h)^2 - 4 = 4-4h+h^2-4 = -4h+h^2 now divide that by h and take the limit.
*September 15, 2015*

**Math**

160 = 8 * (16 + (12 - 4) / 2)
*September 15, 2015*

**physics**

just calculate their heights at time t. You want 85-15t-4.9t^2 = 29t-4.9t^2 The graphs shown here might help visualize things: http://www.wolframalpha.com/input/?i=85-15t-4.9t^2+%3D+29t-4.9t^2
*September 15, 2015*

**Math **

How would I write an algebraic expression for the quotient of fifty and 5 more than a number is 10. Would this be correct; 50/(x+5)= 10
*September 15, 2015*

**physics**

clearly, it is (-19.4 - 19.0)km/s since acceleration is change in v over time, it is (-19.4 - 19.0)km/s ------------------- 1.85 yr Of course, that gives a value in units of km/s-yr, probably not one you want to wind up with. Convert yr to seconds, if you want km/s^2
*September 15, 2015*

**Math**

Hard to say. The sum of the numbers from 1 to 6 is 21. I think you need to be a bit more specific on the details of the figure.
*September 15, 2015*

**Physics**

Draw a diagram. If the swimmer swims upstream at an angle of θ, it is clear that sin θ = 0.50/0.65 Doesn't matter how wide the stream is, does it?
*September 15, 2015*

**MATH**

Well, just split things up till you have 5 terms instead of 2. You do know that terms are separated by + and - signs, right? Here's one way, starting by splitting each of those two terms into two more: 5m^3 + 4n = 2m^3 + 3m^3 + 2n + 2n That's four terms so far. Now ...
*September 15, 2015*

**Math**

Now you are correct.
*September 15, 2015*

**Algebra**

(5+3+4)/3 = 4
*September 15, 2015*

**college algebra**

.314/7.1 = 0.044 = 4.4%
*September 15, 2015*

**PLEASE- physics**

it takes 1.107 s to fall 6m. So, the horizontal velocity must be at least 10m/1.107s = 9.037 m/s I think you can now do the rest. Figure how long it takes to fall the remaining 4 meters to the bottom of the canyon. Then use that to find the horizontal distance moved.
*September 15, 2015*

**Algebra**

-10+(-2)(-2)(-2)/(-3) = -10+(4)(-2)/(-3) = -10+(-8)/(-3) = -10+(8/3) = -22/3
*September 15, 2015*

**Algebra**

#1 no question #2 is almost right. What about the alternating +/- signs?
*September 15, 2015*

**math**

A: No, if y<0 B: No, same as A C: No, consider (1,3) D: No, same as C E: No. Clearly if x < y^2, 3x<3y^2 None of the above
*September 15, 2015*

**Geometry**

If the circumference (and hence the radius) grows by a factor of 2, the volume grows by a factor of 2^3 = 8.
*September 15, 2015*

**math**

a^3+a^2b+a+b = (a+b)(a^2+1) 9a^3+9a = 9a(a^2+1) So, the first factor is (a+b)/9a 8a^2-8b^2 = 8(a-b)(a+b) Now multiply and you have (a+b)(16a^2) ---------------------- (9a)(8)(a+b)(a-b) = 2a / 9(a-b)
*September 15, 2015*

**math**

8x^3+729 is the sum of two cubes, so it is (2x+9)((2x)^2-(2x)(9)+9^2) = (2x+9)(4x^2-18x+81) That help?
*September 15, 2015*

**Math**

clearly the number was between 311,111 and 314,999 and was rounded to the nearest 10,000 It could not be less than 310,000, because then it would have been something like 306,827 with a zero.
*September 15, 2015*

**Knox fields.**

I suspect a typo, as 1m is the same as 1^2 m
*September 15, 2015*

**physics**

mg(h2-h1) = 50*9.8*500
*September 15, 2015*

**Chemistry**

If your reaction is 4Fe + 3O2 -> 2Fe2O3 then it takes 2 moles of Fe to produce 1 mole of rust. 5.0g of rust is 0.0313 moles So, how many grams of Fe in 0.626 moles?
*September 15, 2015*

**algebra**

Enter your numbers here http://davidmlane.com/hyperstat/z_table.html to get a feel for how this stuff works.
*September 15, 2015*

**algebra**

I think this site will be a big help. http://davidmlane.com/hyperstat/z_table.html It will show you the fraction more than 2 std away, and you can multiply that by 15000 to get your answer.
*September 15, 2015*

**Uniport ( physis)**

h + vt - 4.9t^2 = 0
*September 15, 2015*

**Math**

since time = distance/speed, 55/(20-x) = 90/(20+x)
*September 15, 2015*

**Geometry**

Hmmm. Let's see. r^2 = 5^2+5^2 = 50 The area is πr^2, or 50*3.14159 = 157.09 Looks like C to me. If you try stuff and it seems wrong, you really should post your work, so we can help fix it.
*September 15, 2015*

**Geometry**

The circle is (x-8)^2 + (y-2)^2 = r^2 Plug in your point (3,7) and you have (3-8)^2 + (7-2)^2 = r^2 Now that you have r^2, you know the area is just pi r^2
*September 15, 2015*

**geometry**

the length of the diameter is √((7-1)^2+(-6-2)^2) = 10 Now you can find the circumference, right?
*September 15, 2015*

**geometry**

Do you not read what I said? Find the lengths of two of the sides. If the points really do describe a rectangle, one side will be the length, and the other will be the width. I assume you can find the perimeter from that information!
*September 15, 2015*

**geometry**

find the lengths WX and XY. Twice their sum is the perimeter. Use the distance formula: WX = √((-5-(-3))^2+(4-7)^2) = √(2^2+3^2) = √13 and so on
*September 15, 2015*

**math**

There are 3x boys and 2x girls 3x+2x = 40 Now find x, and thus the boys and girls
*September 15, 2015*

**physics**

is there a question in there somewhere?
*September 15, 2015*

**Algebra**

Let the amounts at 1,6,8% be x,y,z. We have: x+y+z = 5000 .01x + .06y + .08z = 310 x = y Now just solve for x,y,z
*September 15, 2015*

**Math 104**

the 2" wheel turns twice as fast as the 4" wheel, since it has half the circumference.
*September 15, 2015*

**physics**

how long does it take to fall 1.1m? Double that for the hang time, since he also has to rise that same distance.
*September 14, 2015*

**Geometry**

well, just do the rule: (x,y) -> (x-4,y+6), so (-7,-3) -> (-7-4,-3+6) = (-11,3)
*September 14, 2015*

**Math**

division by zero is undefined. Everywhere else, the expression is defined.
*September 14, 2015*

**Math 10**

clearly, a circle of diameter 20 cm
*September 14, 2015*

**Calculus**

(f(x+h) - f(x))/h = ((x+h)-x)/h = h/h = 1 no need even to take the limit, since 1 is constant.
*September 14, 2015*

**Geometry**

Review your length formula. Find the lengths of the sides, and add them up. I'll give you one side, you do the other two. AB = √((5-(-2))^2+(-3-2)^2) = √(7^2+5^2) = √74
*September 14, 2015*

**Math**

well, since P(red)=1/2, you need red/1260 = 1/2
*September 14, 2015*

**Precalculus**

2500 * 2*2/14 rev/min * 7/3 * 12 * π in/rev * 60min/hr * 1mi/(12*5280)in = 59.5 mi/hr
*September 14, 2015*

**math**

If x at 6%, then .06x + .09(4x+29000) = 29910
*September 14, 2015*

**Math**

just keep multiplying by -1/2
*September 14, 2015*

**Geometry**

Did you find <2? If so, then read my last line above. <1 and <2 are supplementary.
*September 14, 2015*

**Geometry**

<2 = x^2-3x <4 = 6x+10 <2 = <4, so x^2-3x = 6x+10 x^2-9x-10 = 0 (x-10)(x+1) = 0 x = 10 or -1 Now you can figure out the angles. Remember that <1+<2 = <3+<4 = 180
*September 14, 2015*

**Math**

S = L(1-r) S/L = 1-r I expect you can now find r...
*September 14, 2015*

**math**

x = 280 + .04y Your equation looks better, but it does not meet the explanation of x and y. Maybe there's a typo.
*September 14, 2015*

**Pre-Cal**

looks good to me.
*September 14, 2015*

**Physics**

.13km / (62.9-48.9)km/hr = 0.00928 hr now make that into seconds
*September 14, 2015*

**geometry**

Draw the diagram of what is given. Clearly the garden is 3x7. So, the dimensions of the entire area of garden+path is 9x13. The area of the path is thus 9*13-3*7
*September 14, 2015*

**5th grade math**

(3x9)+(18+36)/9=33
*September 14, 2015*

**Math**

I would naturally expect it to have the value -4(0)-6
*September 14, 2015*

**Calculus**

√(81x^2 + x)− 9x = (√(81x^2 + x)− 9x)(√(81x^2 + x)+ 9x)/(√(81x^2 + x)+ 9x) = ((81x^2+x)-81x^2)/(√(81x^2 + x)+ 9x) = x/(√(81x^2 + x)+ 9x) = 1/(√(81 + 1/x)+ 9) as x->∞, that becomes 1/(√81+9) = 1/18
*September 14, 2015*

**Calculus**

well, what is t when h=0? h(t) = t(19-1.86t)
*September 14, 2015*

**algebra**

(2/3)(3)+(1/4)(10)+(1/3)(20) ------------------------------- 3+10+20
*September 14, 2015*

**Math**

is there a question in there somewhere?
*September 14, 2015*

**math**

98654433445*876689956e347 = 9.8654433445e10 * 8.76689956e355 = 8.6489... * 10^366
*September 14, 2015*

**Mathematics in science**

#1. only 1 significant digit. Trailing zeros do not count. See http://www.rpi.edu/dept/phys/Dept2/yesterday/APPhys1/sigfigs/sigfig/node152.html #2. is correct
*September 14, 2015*

**math**

4m doesn't matter how deep the water is.
*September 14, 2015*

**Maths**

just multiply the tops and bottoms: x y^2 z^2 ----------------- yz z y add the powers of each variable: x y^2 z^2 ------------ y^2 z^2 Now cancel like factors, and you wind up with just x
*September 14, 2015*

**Calculus 1**

√(x+h)-√x ------------ h (√(x+h)-√x)(√(x+h)+√x) -------------------------- h(√(x+h)+√x) (x+h)-x ------------------- h(√(x+h)+√x) h --------------------- h(√(x+h)+√x) The h's cancel, and you have 1/(&#...
*September 14, 2015*

**Calculus 1**

just use the difference quotient, and recall how to "rationalize" expressions involving roots: √x - a = (√x-a)(√x+a)/(√x+a)
*September 14, 2015*

**Calculus 1**

I guarantee that your text has this very problem in it. It starts out as the limit of √(x+h)-√x ------------ h That's hard to handle, but if you multiply top and bottom by √(x+h)+√x things will drop out nicely when you take the limit.
*September 14, 2015*

**Calculus 1**

The slope of the tangent for any x is f'(x) = 2x-1 So, you want the line y-f(3) = f'(3) * (x-3)
*September 14, 2015*

**Calculus 1**

x^2-3x+10 is never zero, so there are no VAs. You found the roots, where y=0.
*September 14, 2015*

**Calculus 1**

c'mon, guy, this is just Algebra II. vertical asymptotes where the denominator is zero and the numerator is not zero. Horizontal asymptotes where x is huge. When x is huge huge huge, only the highest powers matter. So, toss out everything else in the top and bottom and see...
*September 14, 2015*