Friday

December 19, 2014

December 19, 2014

Total # Posts: 27,628

**Calculus I (PLEASE HELP!)**

so, what is confusing you? Newton's method says that, if you have an estimate a root of f(x)=0, you can get a better estimate by using the formula f(x) -> f(x) - f(x)/f'(x) You have the function f(x) = x^2-5, so that f(√5) = 0. f'(x) = 2x. So, you start ...
*November 16, 2014*

**MATH**

It would help if you showed your work. How do you get 5? x is °C f(x) is ml, so the rate of change of ml/°C is just Δf/Δx = 5/5 = 1 For problems like this, you must read the question carefully to see just what they are asking. You gave the quantity of change...
*November 16, 2014*

**calc**

no idea what the regions are. But if f(x) = F'(x), then the areas are x^3+2x+F(x)[2,4] = (64+8+F(4))-(8+4+F(2)) Presumably you have some way of finding F(x)
*November 16, 2014*

**1210 Math**

Assuming you mean f(x) = -x^2 + 2x + 2 you want to find x where f(x) = 0. That is where it hits the ground. So, just solve the quadratic as usual.
*November 16, 2014*

**Calc**

just as the derivative of e^x is e^x, so is the integral. So, you have 2(e^5 - e^-5)
*November 16, 2014*

**linear inequalities**

I see an equation, not an inequity -- nor even an inequality. The equation describes a line.
*November 15, 2014*

**Math (Algebra 1)**

an average of 90 on 4 exams means 4*90=360 points in all. So, how many points shy of 360 is the sum of the first three tests?
*November 15, 2014*

**Calculus**

the integral is 0 if the integration limits are -c and +c, since f(-c) = -f(c) That is not the case here.
*November 15, 2014*

**Calculus**

let u = a^2-x^2 and you have du = -2x dx So the integral becomes ∫[a,0] -1/2 √u du = -1/3 u^(3/2) [a^2,0] = 1/3 a^3
*November 15, 2014*

**Math**

you want permutations of LEER, which is 4!/2! = 24/2 = 12 You have to divide by 2! because the two E's are indistinguishable. LE1E2R looks just like LE2E1R Since there are 2! ways to shuffle the E's, you need to divide the total by 2!. Read up on permutations with ...
*November 15, 2014*

**Algebra: Steve**

c'mon, it's just a quadratic: x(2x^2+5x-12) = 0 x(2x-3)(x+4) = 0 ...
*November 15, 2014*

**algebra**

gas pay = $x/hr tutor pay = $2x/hr sister pay = $0.50x/hr x/2 just means 1/2 x, or 0.5x Not knowing how many hours they work, I can't assign a number. Also, how is the guy's time split between gas jockey and tutor?
*November 15, 2014*

**Math**

wheel radius=22 means the amplitude is 22. So, y is something like y = A+22sin(x) Since the bottom is at y=1, that means y = 23 + 22sin(x) Oops. The wheel starts at the top when t=0, so we really have y = 23 + 22cos(kx) The period is 14, so 2pi/k = 14 k = pi/7 y = 23 + 22cos(...
*November 15, 2014*

**Calculus**

I think you read "nine times" as "nine tenths". Also, the sum of the perimeters is x. In that case, the two perimeters are 9/10 x and 1/10 x. So, the areas sum to ((9/10 x)/4)^2 + ((1/10 x)/4)^2 = 41/800 x^2
*November 15, 2014*

**Advanced Algebra**

We all know that V(r) = 4pi/3 r^3 clearly, f(t) = 21-4.5t V(f(t)) = 4pi/3 (21-4.5t)^3 Now just plug in your numbers
*November 15, 2014*

**Science- Very Urgent**

Of course not. The Bohr model will always be the Bohr model. However, it is always possible that new data will give rise to a new model, possibly based in the Bohr model. Or any other model. Einstein's special relativity did not change Newtonian mechanics in any way. It ...
*November 14, 2014*

**math**

you want to get x by itself, so divide by its coefficient: #2: -3x = 21 divide both sides by -3 and you have x = -7 #5: divide by 2/3 -- that's the same as multiplying by 3/2, so x = 8 * 3/2 = 12 Now you can probably do the others, no?
*November 14, 2014*

**Math**

5lnx-6ln(x+2) ln x^5 - ln (x-2)^6 ln [x^5/(x-2)^6]
*November 14, 2014*

**Math Help Respond To @Steve**

Apparently so, since that is what you provide. My answer is arrived at in this wise: 1000e^0.276t = 6000 e^0.276t = 6 0.276t = ln6 t = ln6/0.276 t = 6.49188 So, 6.5 is the correct answer. Good job.
*November 14, 2014*

**Calculus**

a = 75 km/hr/s = 20.833 m/s^2 s = vt + 1/2 at^2 = 4.1667(11) + (1/2)(20.833)(11^2) = 1306.23 m
*November 14, 2014*

**Algebra**

clearly, f(0) = 0 so, f(x) = ax^3+bx^2+cx = x(ax^2+bx+c) -1(a-b+c) = 15 1(a+b+c) = -5 2(4a+2b+c) = 12 solve those and you wind up with f(x) = 2x^3+5x^2-12x See http://www.wolframalpha.com/input/?i=table+2x^3%2B5x^2-12x+for+x+%3D+-1..2 Now just solve for the roots of f(x)
*November 14, 2014*

**Algebra**

go to calc101.com and click the long division button. Enter your polynomials and see the amazing details!!
*November 14, 2014*

**math**

no inverse. Hopwever, if you meant f(x) = -2x+7, then f-1(x) = (7-x)/2
*November 14, 2014*

**Coordinates Help asap?**

reflection in x-axis takes (x,y) -> (x,-y) you are correct.
*November 14, 2014*

**Algebra**

(-4x^2)(-7) + (2x)(tx) + (-5)(x^2) = x^2(28+2t-5) so, you need 28+2t-5 = 0
*November 14, 2014*

**MATH**

√63 = √(3^2*7) = √(3^2*7^2/7) = 3*7/√7 = 21/√7
*November 14, 2014*

**math**

4^-2 = 1/4^2 = 1/16 2^0 = 1 So, 1/16 * 1 = 1/16
*November 14, 2014*

**Algebra**

First, note that the degree of f must be 3. If higher, there are many possible polynomials which will fit the four points. If f(x) = ax^3+bx^2+cx+d d = 47 a+b+c+d = 32 8a+4b+2c+d = -13 27a+9b+3c+d = 16 a+b+c+d = 32
*November 14, 2014*

**Algebra**

correct
*November 14, 2014*

**math**

√24 = √4√6 = 2√6 There are no other perfect squares which are factors
*November 14, 2014*

**geomerty**

use the distance formula. So, d = √((3-0)^2 + (-5-3)^2)
*November 14, 2014*

**Math Help**

just solve for t in 1000e^0.276t = 6000 e^.276t = 6 .276t = ln6 t = ln6/0.276
*November 14, 2014*

**algebra**

|g-24995| < 1000 That is, if exactly $1000 off loses.
*November 14, 2014*

**Math**

sinθ = 8/17 in QII means that cosθ = -15/17 sin2θ = 2sinθcosθ = 2(8/17)(-15/17) cos2θ = cos^2θ - sin^2θ = (15/16)^2 - (8/17)^2 tan2θ is of course, sin2θ/cos2θ
*November 14, 2014*

**Astronomy **

How long does it take a protostar with one solar mass to become a main sequence star? 1 or 3 or 10 or 30 or 100 million years?
*November 14, 2014*

**MATH**

(3,7): 2*3+2*7 = 20 and so on
*November 14, 2014*

**Algebra**

If you mean √(2x-4) = √(7x+9) then it's easy. Squaring both sides gets rid of the roots, and you have 2x-4 = 7x+9 5x = -13 x = -13/5 But, then 2x-4 < 0 and √(2x-4) is undefined. So, since you say that "isn't working" I assume you meant &#...
*November 14, 2014*

**sdfgsdfgsd**

IYuyuiyuiyui !!
*November 14, 2014*

**Math**

Consider each line as if it were an equation. For example, 3x+4y = 8 That describes a line in the x-y plane. But, you have an inequality. So, the solution is the entire half-plane below the line, where 3x+4y <= 8 Same for the other inequality. So, the solution set for the ...
*November 14, 2014*

**English/grammar**

1. Whose is this wallet that was found in the room yesterday? 2. A man with five children has a large family to support.
*November 14, 2014*

**Calculus**

f(x) = √(x^3+4x) f'(x) = (3x^2+4) / 2√(x^3+4x) f'(2) = 16/8 = 2 So, now you have a point and a slope. The line is y-4 = 2(x-2) y = 2x See the graphs at http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%28x^3%2B4x%29%2C+y%3D2x+for+x%3D-1..3
*November 14, 2014*

**math**

32/100 That can, of course, be reduced as desired.
*November 14, 2014*

**sixth grade math**

do you mean 24% ? If so, then 360 = .24x x = 360/.24 = 1500
*November 14, 2014*

**sixth grade math**

I'm having trouble with the English: What is the "of" doing there?
*November 13, 2014*

**Calculus Please Help**

F(x,y) = e^(2x+y) Fx = 2e^(2x+y) Fy = e^(2x+y) Fx(1,-1) = 2e^(2-1) = 2e Fy(1,-1) = e^(2-1) = e F(x, y, z) = xy + xz -yz Fx = y+z Fy = x-z Fz = x-y Fx(0,-1,1) = 0 Fy(0,-1,1) = -1 Fz(0,-1,1) = 1 F(x,y,z) = 2x^0.2 y^0.8 + z^2 Fx = 0.4x^-0.8 y^0.8 Fy = 1.6x^0.2 y^-0.2 Fz = 2z Fx(0...
*November 13, 2014*

**Math**

multiply top and bottom by -9-9i, Note that (-9+9i)(-9-9i) = 9^2+9^2 = 162 So, you have (-10-6i)(-9-9i)/162 Now just do the multiplication and put into standard form.
*November 13, 2014*

**Trigonometry**

well, just plug in the numbers: cot x = (3.5)(9.8)/5.4^2
*November 13, 2014*

**Algebra**

(a) f and g are both degree 4, so their sum cannot be of higher degree. (b) 1/2 g(x) = x^4-3x^2+x - 1/2 f - 1/2 g = -x + 5/2 Looks like 1 is the smallest possible degree.
*November 13, 2014*

**Math**

You might take all 6 red balls, all 5 green balls first. So, you might have to take out 6+5+3=14 balls to be sure you also have 3 yellows. Extra credit: how many do you need to be sure you have 3 balls of some color? Hint: There are 3 colors.
*November 13, 2014*

**Math**

Thanks alot Reiny that was super helpful. I have a math test tomrrow!
*November 13, 2014*

**Math**

I'm sorry but this does not help. WHat is D?
*November 13, 2014*

**Math**

Can anyone show me how to solve this? .1g+3.75m=7.36 and 8.2g+21m=73 Thank you
*November 13, 2014*

**Math help please!!!!!!**

correct again
*November 13, 2014*

**Math (PLEASE!)**

But you are right. D is also true.
*November 13, 2014*

**Math (PLEASE!)**

well, 11 is greater than 10.7 11 > 10.7 11.1 is even greater 11.1 > 11 > 10.7 Can you use the number line? If, plot all those values. The ones on the right are greater than those to their left.
*November 13, 2014*

**Math (PLEASE!)**

I like C
*November 13, 2014*

**Math help please!!!!!!**

correct
*November 13, 2014*

**Algebra**

I get -15 6x^3*4 + 4x^2*3x + -7x*3x^2 + -5*6x^3 24+12-21-30 = -15
*November 13, 2014*

**Algebra**

h(x) = (3x^2-5x+3)-(9x^3-3x+1) so, just subtract tghe coefficients of like powers.
*November 13, 2014*

**math take 2**

fcator out 4tan^2θ to get 4tan^2θ(2sinθ+1) = 0 so, tanθ = 0 or sinθ = -1/2 θ=0 or θ = 7π/6 or θ = 11π/6 You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be &#...
*November 13, 2014*

**math**

fcator out 4tan^2θ to get 4tan^2θ(2sinθ+1) = 0 so, tanθ = 0 or sinθ = -1/2 θ=0 or θ = -π/6 You didn't specify the domain, so I'll let you worry about that. If you want all solutions, then that would be θ = πn θ...
*November 13, 2014*

**science**

the same
*November 13, 2014*

**Math *checking answers*?**

Nope - my bad. B,C,D are all the same You want 9.5/3 = x/7 but B,C,D are all 9.5/3 = 7/x
*November 13, 2014*

**Math *checking answers*?**

I like C for #2. It's the same as 9.5/3 = x/7
*November 13, 2014*

**calc**

I think you want g(x) = 3x√(x+5) on [-5,0] g(-5) = 0 g(0) = 0 So, we want c somewhere in (-5,0) such that f'(c) = 0 f'(x) = 3(3x+10) / 2√(x+5) f' = 0 when 3x+10 = 0 So, c = -10/3 which is in the domain.
*November 13, 2014*

**calc**

In that case, g(0) = 0 g(5) = 15√10 Again Rolle's Theorem does not apply. Try again.
*November 13, 2014*

**calc - eh?**

g(0) is not defined, since sqrt(-5) is not real. So, g(x) does not satisfy the condition that g(0) = 0 You want to reconsider any part of the question?
*November 13, 2014*

**MATH HELP PLEASE**

(13/4)(2x + 3/4) (13/4)(2x) + 13/4 * 3/4 (13/2)x + 39/16
*November 13, 2014*

**Math: Chi Squared**

Boy, you don't give much context, do you?
*November 13, 2014*

**Intermediate Algebra**

clearly the domain is not (-oo,100) negative percent makes no sense. I'd say the domain is [0,100) C(X) -> oo as x->100, meaning that the harder you try to squeeze out that last bit of pollution, the morse it costs. So, just evaluate C(90). Then double that, since we...
*November 13, 2014*

**Algebra**

-3x^5 + 24x^2 factor out -3x^2 and you have (-3x^2)(x^3) + (-3x^2)(-8) -3x^2 (x^3-8) Now recall the factorization of the difference of two cubes, and you have -3x^2(x-2)(x^2+2x+4)
*November 13, 2014*

**Science**

F = ma so, a = F/m the x-component is F cos60° and sine for y
*November 13, 2014*

**Molecular Calculations**

convert g of Sn to moles. You will need that many moles of O2. Then find the mass for that many moles of O2.
*November 13, 2014*

**Algebra 2**

7x^2 = 21x 7x^2-21x = 0 7x(x-3) = 0 This is why we always set stuff to zero. If the product of factors is zero, one of the factors must be zero. So, we have 7=0: never x=0: x=0 x-3 = 0: x=3 Those are the solutions: x = 0 or 3
*November 13, 2014*

**math**

0.8 = 8/9 So, you have 10.1 + 8/90 = 101/10 + 8/90 = 917/90
*November 13, 2014*

**Calculus**

yes
*November 12, 2014*

**math**

at any depth y, the radius of the surface of the water is y/2. So, v = pi/3 (y/2)^2 y = pi/12 y^3 dv/dt = pi/4 y^2 dy/dt Now plug in your numbers.
*November 12, 2014*

**Calculus - PLEASE HELP!!**

Looks good to me
*November 12, 2014*

**math**

(6.0-0.3)/4
*November 12, 2014*

**Inverses of functions**

f^-1(a-1) = -3 a-1 = f(-3) a-1 = 2 a = 3 Not sure just what you're asking
*November 12, 2014*

**math**

b/g = 8/5 b = g+135 now just solve for g
*November 12, 2014*

**Pre-Calculus**

Look what happens when x gets huge. The equation is basically 5y^2 = 20x^2 y^2 = 4x^2 so the slopes are 2 and -2
*November 12, 2014*

**Math**

x-25.6 = 84.3
*November 12, 2014*

**Math**

m-4=11 m+u=53 now just find u
*November 12, 2014*

**Math**

(100-50)/(3-1.5)
*November 12, 2014*

**Science**

you are correct
*November 12, 2014*

**math**

s=m m = 4(s-18) Each had 24
*November 12, 2014*

**Pre-Calc/Trig**

range is all positive >2, as well as all negative On the interval (-1,1) the numerator is positive and the denominator is negative.
*November 12, 2014*

**Math HELP!**

x-10 = 47 x-25.6 = 84.3
*November 12, 2014*

**Algebra HELP!**

If the population is P, then 14/50 * P = 30
*November 12, 2014*

**Math - eh?**

eh? try that in English
*November 12, 2014*

**Arithmetic Progression**

a + a+d = 4 a+9d = 19 a=1 d=2 So, now add T5 and T6
*November 12, 2014*

**Algebra**

the two roots have product c and sum 7 Since the coefficient of x is 1, all the rational roots are integers. (x-1)(x-6) = x^2-7x+6 (x-2)(x-5) = x^2-7x+10 ... I think you can see that there are only a few values for c.
*November 12, 2014*

**Calculus**

∫v(v^3+5)^2 dv = ∫v(v^6 + 10v^3 + 25) dv = ∫v^7 + 10v^4 + 25v Now just integrate each term.
*November 12, 2014*

**caculus**

f"(x) = -4sin(2x) f'(x) = 2cos(2x) + c1 f(x) = sin(2x) + c1*x + c2 Now plug in the initial conditions to find c1 and c2
*November 12, 2014*

**calculus**

a(t) = 36t+10 v(t) = 18t^2+10t+c v(0)=9, so c=9 and v(t) = 18t^2+10t+9 s(t) = 6t^3+5t^2+9t+c s(0) = 16, so c=16 and s(t) = 6t^3+5t^2+9t+16 So, find s(11)
*November 12, 2014*

**Calculus**

Try this site: http://mathworld.wolfram.com/RiemannSum.html
*November 12, 2014*

**alg 2**

think back to algebra I. The QF is x = [-b +/- sqrt(b^2-4a)]/2a So, just plug in your number. The first is x = [-4 +/- sqrt(16-4(1)(-2)]/2 = [-4 +/- sqrt(16+8)]/2 = -4 +/- sqrt(24)]/2 = -2 +/- sqrt(6) Do the same with the others. First step is to get everything on the left ...
*November 12, 2014*

**Algebra 2 HELP plz**

t=2 is the correct time. So, plug in t=2 for your formula for h to get the max height. to see when it hits the ground, just set h=0: -16t^2+64t+80 = 0 -16(t+1)(t-5) = 0 so, h=0 at t=5
*November 12, 2014*

**Pre-Calc/Trig...**

2(x+5)/(x-3)
*November 12, 2014*

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