Friday

May 6, 2016
Total # Posts: 275

**algebra**

Nurse erika needs to prepare 12 liters of a 10% saline solution. The stockroom only has 7 % and 15% saline solutions. How much of each solution should erika mix together to get 12 litres of a 10% saline solution
*October 21, 2012*

**algebra**

How many ounces of a 19% copper alloy metal must be added to 42% ounces of a 40% copper alloy metal to produce a metal which is 31% copper alloy
*October 21, 2012*

**algebra**

Nurse erika needs to prepare 12 liters of a 10% saline solution. The stockroom only has 7 % and 15% saline solutions. How much of each solution should erika mix together to get 12 litres of a 10% saline solution
*October 21, 2012*

**Chemistry**

It's 60 C for the reason of its must be higher then forty
*February 23, 2011*

**3rd grade math - arrays**

11 > 10. But 11 can be only 1x11 or 11x1 matrix (array) But 10 can be 2x5, 5x2, 10x1, 1x10.. so mary is wrong to say.... same goes with any Prime number as it can be only 2 ways to sent that count in an array.
*February 9, 2011*

**Calculus**

Thank you!
*December 9, 2010*

**Calculus**

Find the derivative of y=5^(1-4x) lny=(1-4x)ln5 Can I split it? lny=ln5-4xln5 1/y dy/dx = -[4ln5/(4xln5)^2] dy/dx=-[4ln5/(4xln5)^2]y Thanks.
*December 9, 2010*

**Calculus**

I need help to start this problem please, A company can sell 4500 pairs of sunglases monthly when the price is $5. When the price of a pair of sunglasses is increased by 10%, the demand drops to 4250 pairs a month. Assume that the demand equation is linear. Find the demand ...
*December 9, 2010*

**Calculus**

I did the derivative wrong! Thanks!
*December 8, 2010*

**Calculus**

To construct a tincan, V=32pi m^3, The cost per square meter of the side is half of the top and bottom of can. What are the dimensions and the cost? V=πr²h=32pi SA=2πr²+2πrh h=32/r² Domain={r>o} Let x be the cost, I subbed 32/r² for h C=f(...
*December 8, 2010*

**Calculus**

Find the horizontal asymptote of f(x)=e^x - x lim x->infinity (e^x)-x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0-infinity = - infinity lim x-> -infinity (e^x)-x= infinity
*December 8, 2010*

**Calculus**

So the whole interval is increasing? The function of cosx has a section of decreasing though.
*December 5, 2010*

**Calculus**

Wow, so confusing. Thanks!
*December 5, 2010*

**Calculus**

Find the interval(s) where the function is increasing where and it is decreasing. f(x)=sin(x+(pi/2)) for 0≤x≤2pi so my derivative is f'(x)=cos(x+(pi/2))? and my critical number is -pi/2?
*December 5, 2010*

**Calculus**

Shouldn't it be -2xlog(base2)x because the product rule is inside the bracket, and its subtracting the bracket. y"=[x/ln - 2xlog(base2)x -2x²/xln2 - 4x/ln2 + 8xlog(base2)x]/x^5 =[-3x/x^5] + [6xlog(base2)x/x^5] - [2x²/xln2] Then I put them in common ...
*October 31, 2010*

**Calculus**

Find the second derivative of the following function: y=log(base2)x/x² y'=[(x/ln2)-2xlog(base2)x]/x^4 the derivative of x/ln2 is (ln2)^-1? y"={[((ln2)^(-1))-(2log(base2)x/xln2)]x^4-4x³[(x/ln2)-2xlog(base2)x]}/x^8 I tried simplifying it and I got, y"=[3...
*October 31, 2010*

**Calculus**

I have the same thing now, thanks! I thought I had to use product rule for xln2, but there's no variable in ln2, so it is accounted for a constant?
*October 30, 2010*

**Calculus**

I have trouble finding the second derivative, my answer is a little off from the answer key. f(x)=(log(base2)x)^5 f'(x)=[5(log(base2)x)^4]/xln2 f"(x)=[20(log(base2)x)³(xln2)-(ln2+(x/2))(5(log(base2)x)^4)]/[x²(ln2)²] for the f"(x), is g(x)=xln2? It ...
*October 30, 2010*

**Calculus**

So the negative is not included, for example f(x)=-2^x f'(x) would be (-2^x)(ln2)? Thanks!
*October 29, 2010*

**Calculus**

Find the derivative of f(x)=x(1-4^x) f'(x)=(1-4^x)+ (-4^x)(ln-4)(-4)(x) ln can't be negative. Is the derivative correct?
*October 29, 2010*

**Calculus**

Find the linear approximation L(x) of the given function at a, and use it to approximate the function value at the given x-value. f(x)=x^(1/3), a=-27, x=-28 I found the derivative to be f'(x)= 1/[3(-27)^(2/3)] but it does not exist because a number cannot be negative in a ...
*October 26, 2010*

**math**

When I factor out the bottom its x=-3 and x=2, how did you get -2 and 1? Why is +2 included? Doesn't it become undefined when x=2?
*October 5, 2010*

**math**

What's the domain of √((x-2)/(x^2+x-6)) (2,infinity] ?
*October 5, 2010*

**Calculus**

Thank you so much! :)
*September 21, 2010*

**Calculus**

Is the answer 47/9?
*September 21, 2010*

**Calculus**

lim x->1 (x^47-1)/(x^9-1) With what method do I factor them with? Pascal triangle?
*September 21, 2010*

**11th grade**

While driving his sports car at 20.0 down a four lane highway Eddie comes up behind a slow-moving dump truck abd decides to pass it in the left lane. If Eddie can accelerate at 5.00 m/s^2, how long will it take for him to reach a speed of 30.0 m/s?
*September 1, 2010*

**math**

What is 4.38E+01 equal to?
*June 18, 2010*

**Chemistry**

Consider the following buffer equilibrium: HF (high concentration) + H2O <--> H3O+ (low concentration) + F- ( high concentration) Using Le Chatelier's Principle, explain what happens to the pH of the buffer solution when a small amount of NaOH is added. Wouldn't ...
*June 3, 2010*

**Calculus**

Find the area between y=cosx and y=sinx from 0 to 2pi. To find the zeros, I combined the equations cosx-sinx=0 What's next?
*May 21, 2010*

**Chemistry**

Thank you so much for the clarification!
*May 4, 2010*

**Chemistry**

Oh, so they're all bases. Is the answer C because it is the strongest base out of the four. Does the stronger the base mean the more reaction? Thanks.
*May 4, 2010*

**Chemistry**

Water will react most completely as an acid with A. SO3^-2 B. H2BO3- C. C6H5O- D. CH3COO- Why isn't the answer A, since H2O + SO3^-2 <-> HSO3- + OH- so H2O and HSO3- are the conjugate acids.
*May 4, 2010*

**Chemistry**

What is the mass of NaOH required to prepare 100.0mL of NaOH(aq) that has a pH = 13.62 ? A. 0.38g B. 0.42g C. 1.67g D. 2.40x10^-14 g OH- + H2O <-> H3O+ + O^-2 antilog-13.62 = 0.09976 #gNaOH=(0.1-0.09976)M x 40g/mol x 0.1L = 0.40g?
*May 4, 2010*

**Chemistry**

Alright, thanks! It's the correct answer on the answer key.
*May 4, 2010*

**Chemistry**

S^-2 ion has equilibrium constant of 7.7x10^-1 So it's just (1.00x10^-14)/(0.77)=1.3x10^14 = Ka HS^- ?
*May 4, 2010*

**Chemistry**

The S^2- ion is a relatively strong base with an equilibrium constant of 7.7x10^-1. What is the Ka value for HS- ? I got the equation HS^- <-> H^+ + S^-2. Am I able to find [S^-2] with the Keq?
*May 4, 2010*

**Chemistry**

I understand now, thanks for your help! :)
*May 3, 2010*

**Chemistry**

How do you know if D ionizes or not? Is it because it doesn't have a metal or because it is a weak acid? So the more ions it have, the better the conductivity, eventhough HCl is one of the strongest acids?
*May 3, 2010*

**Chemistry**

Which solution will have the greatest electrical conductivity? A. 0.50MHCl B. 0.10MRbOH C. 0.50MK3PO4 D. 2.0MC6H12O6 I thought the answer would be B, but it is C. Is it not B because the Ka is too small? Otherwise, C is has the smallest Ka.
*May 3, 2010*

**Calculus**

Thanks for your help!
*April 29, 2010*

**Calculus**

Differentiate: y=ln(x/(2x+3))^(1/2) y=(1/2)ln(x/(2x+3)) y'=0+(1/2)((2x+3-2x)/(2x+3)²) What's next after the quotient rule?
*April 29, 2010*

**Calculus**

I understand now, thanks! I changed the log to, 1/xln3 instead.
*April 29, 2010*

**Calculus**

Find the derivative. f(x)=(1+logᴈx)/x f'(x)=[1-ln3-logᴈx (ln3)]/(x²ln3) How do you simplify it further? Thanks.
*April 29, 2010*

**Chemistry**

Thanks!
*April 27, 2010*

**Chemistry**

Write the net ionic equation for the hydrolysis reaction that occurs when Ba(CH3COO)2 is dissolved in water. CH3COO- + H2O <-> CH3COOH + OH- ?
*April 27, 2010*

**Calculus**

at the point where x=0 *
*April 19, 2010*

**Calculus**

Find the equation of the tangent line to the curve y=1+ xe^(2x) y'=(e^(2x))(2x+1) = slope What's the next step?
*April 19, 2010*

**Calculus**

Ohh, so that's how you find dy/dt, thanks!
*April 9, 2010*

**Calculus**

A radar antenna is located on a ship that is 4km from a straight shore. It is rotating at 32rev/min. How fast does the radar beam sweep along the shore when the angle between the beam and the shortest distance to the shore is pi/4? Let's say that the shoreline is y, are we...
*April 9, 2010*

**Calculus**

Evaluate the following limit: lim x->(pi/2) [((pi/2)-(x))/cosx] May I please get a hint on how to start this question?
*April 9, 2010*

**Calculus**

Evaluate the following limit: lim x->0 xcscx =limx->0 x/sinx Does it equal one because the reciprocal is 1/1=1?
*April 9, 2010*

**Calculus**

Thank you for the help!
*April 8, 2010*

**Calculus**

Find dy/dx in the following: y=[x/√(1-x²)]- sin‾¹x y'={[(1-x²)^(1/2)-(-2x)(x)(1/2)(1-x²)^(-1/2)]/(1-x²)} - (1/√(1-x²)) Then I multiplied (-2x)(x)(1/2) together. How do I simplify the rest?
*April 8, 2010*

**Calculus**

I did try to continue with it. I brought sec²θ to the top -> cos²θ, so I plugged in cos(20/25). How come you typed 25/20 instead of sec(25/20)²?
*March 30, 2010*

**Calculus**

I'm not good with drawing diagrams from word problems, so I switched the numbers around to see if I could get the correct answer. 4/sec²θ=dθ/dt cosθ=20/25 4cos²θ/20=dθ/dt dθ/dt=0.160rad/s but the answer is 0.128rad/s.
*March 30, 2010*

**Calculus**

A vehicle moves along a straight path with a speed of 4m/s. A searchlight is located on the ground 20m from the path and is kept focused on the vehicle. At what rate (in rad/s) is the searchlight rotating when the vehicle is 15 from the point on the path cloest to the ...
*March 30, 2010*

**French**

Merci beaucoup!
*March 25, 2010*

**French**

Is this correct? He saw construction workers building new buildings and he thought they worked hard for their dream and the buildings represent hope and future. Il a vu les travailleurs de construction de construire de nouveaux bâtiments, et il pensait les travailler ...
*March 25, 2010*

**Calculus**

Thanks. The answer key says the 12 is positive.
*March 24, 2010*

**Calculus**

Find the derivative of y=cot³(1-2x)² y'=3[cot(1-2x)²]² (-csc²(1-2x)²)(2-4x)(-2) y'=(-6+24x)[cot(1-2x)²]²(-csc²(1-2x)²) How do you simplify further?
*March 24, 2010*

**Calculus**

I got-4sin2x for the second term, I made a typo. How do you translate sinx=-1/4 to decimal numbers?
*March 24, 2010*

**Calculus**

Determine the concavity and find the points of inflection. y=2cosx + sin2x 0≤x≤2pi y'=-2sinx + 2cos2x y"=-2cosx-4sinx How do I find the IP(s)?
*March 24, 2010*

**Calculus**

Got it.
*March 24, 2010*

**Calculus**

Find an equation of the tangent line to the given curve at the given point y=(1/cosx)-2cosx at((pi/3),1) y'=0+2sinx slope = √3 y=mx+b 1=(√3)(pi/3)+b b=(3-√3pi)/3 3√3x-3y+3-√3pi=0 What did I do wrong? Thanks in advance.
*March 24, 2010*

**Calculus**

Oh, I see it now, thanks!
*March 21, 2010*

**Calculus**

Find the derivative of y with respect to x: y=(1+cos²x)^6 y'=6(1+cos²x)^5 How do you derive inside the brackets? The answer says -sin2x, but wouldn't it be -2sinx, using the chain rule.
*March 21, 2010*

**Calculus**

Find the derivative of y with respect to x: y=3sin^4(2-x)^-1 y=[3sin(2-x)^-1]^4 y'=4[3sin(2-x)^-1]^3 (-3cos(2-x)^-1)(-1) -(2-x)^-2 y'=[12cos(2-x)^-1][3sin^3(2-x)^-1][2-x]^2 but the answer does not have a 3 in front of sin. What happened to the 3?
*March 21, 2010*

**Calculus**

Find the derivative of y with the respect to x: y=sin‾²(x³) y'=cos‾²(x³)(3x²) y'=3x²cos‾²(x³) How come my answer is incorrect? Thanks in advance.
*March 21, 2010*

**Chemistry**

What will be the [Cl-] when equal volumes of 0.10MNaCl and 0.20MAlCl3 are combined? A. 0.35M B. 0.15M C. 0.30M D. 0.70M
*March 7, 2010*

**Chemistry**

Is it, Final [Pb 2+]=0.0020M x (100ml/200ml)=0.0010M Pb 2+ Final [I-]=0.0040M x (100ml/200mL)=0.0020M I- Ksp=(0.0010)(0.0020)^2 Ksp=4.0x10^-9
*March 7, 2010*

**Chemistry**

The question asks for the ksp, so don't I try to find the final concentration first? How do you take 100mL of each molar concentration?
*March 7, 2010*

**Chemistry**

In an experiment, a student mixes equal volumes of 0.0020M Pb 2+ ions with 0.0040M I- ions. The trial ion product is? PbI2 <-> Pb 2+ + 2I- How do you determine the final concentration of Pb 2+ and I-?
*March 7, 2010*

**Chemistry**

Thanks! I'll keep that in mind.
*March 5, 2010*

**Chemistry**

Consider the following equilibrium: CaCO3(s)<->Ca 2+(aq) + CO3 2-(aq) Which of the following reagents, when added to the equilibrium system, would cause more CaCO3 to dissolve? a)KNO3(s), can't be this, NO3 is soluble with all negative ions b)CaCO3(s), adding more ...
*March 5, 2010*

**Calculus**

Retyped with brackets, it would be If √2[cosx)-1 = (1+√3)/2 and √2(cosx)+1 = (1-√3)/2, find the value of cos4x.
*March 5, 2010*

**Calculus**

I'm pretty sure I typed the question correctly, the answer at the back says 1/2.
*March 5, 2010*

**Calculus**

If √2cos(x)-1 = (1+√3)/2 and √2cos(x)+1 = (1-√3)/2, find the value of cos4x.
*March 5, 2010*

**Calculus**

Thanks!
*March 5, 2010*

**Calculus**

Find the value of tan2x, (pi/2)<x<pi, given secx=-5/4. So cosx=-4/5? I've no clue what to do next.
*March 5, 2010*

**Physics**

Name three ways in which you can reduce the distortion due to diffraction of a wave passing through an aperture. -large aperture -small wavelengths What's the last one?
*March 2, 2010*

**Calculus**

Got it, thanks!
*February 27, 2010*

**Calculus**

Solve identity, (1-sin2x)/cos2x = cos2x/(1+sin2x) I tried starting from the right side, RS: =(cos²x-sin²x)/(1+2sinxcosx) =(cos²x-(1-cos²x))/(1+2sinxcosx) and the right side just goes in circle. May I get a hint to start off?
*February 27, 2010*

**Calculus**

I did use x^(1/2). I got y'=[1/2x^(-1/2)]/(√x+1)^2, then I brought the top to the bottom. The last part of the question is sketching it. The graph is basically f(x)=√x with a horizontal asymptote at y=1. Thank you Reiny :)
*February 12, 2010*

**Calculus**

So my first derivative is wrong? Is the first step to the second derivative, y"=-(2√x(x+1)^2)^-2 (-1) ?
*February 12, 2010*

**Calculus**

y=(√x)/((√x)+1) Find the concavity and point of inflection(s). y'=1/(√(x/2))(√x+1)² y'=1/(x√(x/2)+2√(x²/2)+√(x/2)) y"=0- ? How do you find the second derivative?
*February 12, 2010*

**Physics**

The wavelength of red light is about 7x10^-7m. The frequency of the red light reflected from a metal surface and the frequency of the vibrating electron that products it are the same. What is this frequency?
*February 12, 2010*

**Calculus**

Then isn't the domain -1≤x≤1, since when I plug in -1 and 1, the answer is 0. For symmetry, I plugged 3 into y=x/[(1-x²)^(-2)], which I'm guessing is wrong. Since domain is -1≤x≤1, I switched the number to 0.5, and found that it is ...
*February 11, 2010*

**Calculus**

y=x√(1-x²) Domain? y=x(1-x²)^(1/2) y=x/[(1-x²)^(-2)]? x≠±1 Symmetry? f(3)=192 f(-3)=-300 -f(-3)=300 no symmetry?
*February 11, 2010*

**Chemistry**

I understand now, thanks for clarifying. :)
*February 9, 2010*

**Chemistry**

Isn't the equation not balanced then?
*February 9, 2010*

**Chemistry**

Write an equation to show the dissolving of the following substance in water. (NH4)2SO4(s) NH42SO4 + H2O -> 2NH4 + SO4 + H^+ + OH-?
*February 9, 2010*

**Physics**

I thought the medium travels horizontally in a wave. Thanks.
*February 9, 2010*

**Physics**

Does the medium in which a wave travels move along with the wave itself. I think the answer is no, but why?
*February 9, 2010*

**Chemistry**

Oh, so Mg(OH)2 only decreases on mole or solid. I got confused because OH- was on both sides. Thanks so much :)
*February 7, 2010*

**Chemistry**

Consider the following equilibrium: Mg(OH)2 (s) <--> Mg 2+ (aq) + 2OH-(aq) What happens to the amount of solid Mg(OH)2 when some HCl is added? It decreases because it has least entropy? So in the reaction, both the concentration of Mg(OH)2 and OH- decreases?
*February 7, 2010*

**Chemistry**

So it is c because the molecules move faster in gas. Thank you!
*February 7, 2010*

**Chemistry**

Which of the following statements correctly describes the change which occurs when a liquid vaporizes at its boiling point at a given external pressure. a) The entropy decreases b) The temp increases c) The kinetic energy increases d) The potential energy increases I can't...
*February 7, 2010*

**Chemistry**

Whoops, forgot the rules to balancing an equation for a second, thanks! :) Is the activity of pure liquids, 1, as well?
*February 1, 2010*

**Chemistry**

There are 3 mols on each side. I made the question up, so I think HCl was supposed to be a gas. I was just wondering if it would be Keq = 0/(Cl2)(HI)^2 or Keq= 1/(Cl2)(HI)^2. Thanks for clearing it up.
*February 1, 2010*