# Posts by Steve

Total # Posts: 51,710

**Math**

I agree

**Maths**

you need to learn to use parentheses online to make it clear what you want, since otherwise, multiplication gets done before addition. You meant (cot^2(x)(sec(x)-1))/(1+sin(x))=(sec^2(1-sin(x))/(1+sec(x)) Apply some of your elementary identities: cot^2(sec-1) = (sec-1)/tan^2...

**Maths**

recall that (a^3-b^3) = (a-b)(a^2+ab+b^2) sin^2+cos^2 = 1 and it all drops right out.

**chemistry**

for the same reason that any other endothermic reaction is so considered...

**Calculus**

there are lots of online graphers. You already know what the graph looks like. The critical points are where secx=0 or tanx = 0 The concavity up or down is provided by the sign of y"...

**MAth**

so, just plug in the numbers, and you should get 129

**College Algebra**

no, it is actually .05x + .10y + .25z = 7.80 Now solve 'em!

**maths steve reiny bob reiny!!! Help**

One side is the distance from (3,4) to 2x+y+3=0 The other side is the distance from (3,4) to x-2y-1=0 The area of the rectangle is thus |2*3+1*4+3)/?(2^2+1^2) * |1*3-2*4-1|/?(1^2+2^2) = 13

**Calculus**

more like dy/(1+y) = x dx ln(1+y) = x^2/2 1+y = c e^(x^2/2) y = c e^(x^2/2) - 1

**Physics**

ever hear of google? http://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement

**Pre calc**

so, why don't you show us how you got it? I get in QIII tan? = sin?/cos? = 5/12, so sin? = -5/13 cos? = -12/13 in QIV sin? = -1/?10 cos? = 3/?10 You don't say what you want to do with that, but sin(???) = (-5/13)(3/?10)-(-12/13)(-1/?10) = 21/(13?10)

**Calculus**

See your other post.

**Math**

a nominal rate of r compounded n times per year is (1+ r/n)^(nt) So, since you have a 4t there, .01 = r/4, making r=4% (1 + .04/4)^(4t)

**Math**

3.3*0.006*10 = 0.198 12*0.005*200 = 12 .198/12 = 0.0165 = 1.65x10^-2

**Math**

ok, so we have a circle and a tangent. A tangent to the circle will always be perpendicular to a line through the center. Now what? Ever hear of a space bar?

**Math**

(3.25*8)x10^(-22+3) = 26x10^-19 = 2.6x10^-18

**pre calc**

what are all those sin() stuff still hanging around? The final expression should just be a bunch of fractions. tan(? + ?); cos(?) = ? 1/3 ? in Quadrant III, sin(?) = 1/4 ? in Quadrant II You really need to review the signs of the trig functions in the various quadrants. In ...

**pre calc**

me either. why don't you show your work? I've done several already; your turn. I'm sure we can pinpoint where you go astray.

**Pre calc**

Funny - you and Ruth need to compare notes ... in QI sin? = 5/13 cos? = 12/13 in QII sin? = 1/?5 cos? = -2/?5 sin(?+?) = (5/13)(-2/?5)+(12/13)(1/?5) = 2/(13?5)

**Pre calc**

Better check my math, then. Also, maybe they don't want the radical in the denominator.

**Pre calc**

in QIII sin? = -5/13 cos? = -12/13 in QIV sin? = -1/?10 cos? = 3/?10 sin(? ? ?) = (-5/13)(3/?10)-(-12/13)(3/?10) = 21/(13?10)

**Math - Calc**

You forgot the chain rule. The result is (x^2)^(x^2) * 2x if f(x)=?[1,u] t^t dt and u = u(x) then f'(x) = f(u) * du/dx

**Math**

well, 141 is 11/12 std above the mean...

**Calculus - Definite Integrals Please Help!**

If you want h'(x), then that would be [cos(sin^3(x))+sin(x)]*cos(x)

**Math - Calculus**

((3x+2)/(3x-5))*3 - ((6x+2)/(6x-5))*6

**statiictis**

So, you want 4 of each kind. Pick any set of 4 from each kind, and then there will be 12! ways to play them. 21C4 * 13C4 * 25C4 * 12! = 2.593*10^19

**Calculus**

((1/2)x^2-1)^8

**math**

It's easier if you rewrite it as u = 1/x^4 u^2-u+1 = u^2-u+1/4 + 3/4 = (u - 1/2)^2 + 3/4 = (1/x^4 - 1/2)^2 + 3/4

**Math**

If all the angles are in QI, then sinA = 4/5 cosA = 3/5 sinB = 12/13 cosB = 5/13 now just use the difference formula

**Science! NEED HELP URGENT**

none. wavelength and velocity are related, but not amplitude. Now, if you're talking about something like a water wave, where water molecules are moving up and down, then of course, amplitude is related to the velocity of the moving particles.

**Geometry**

102lb * 0.45kg/lb * 2.5mg/kg = 114.75 mg

**English**

I think google is very helpful for these idioms. You can get various definitions and explanations.

**Math**

Maybe if you explain what you mean by a road a pound we can work it out. However, the perimeter of the drawing is 16 inches, and 16*5 = 80 so ...

**Language Arts**

ever see a cold, dreary, misty winter day? (between dawn and dusk, both the boundaries of the dark of night)

**trigonomenty**

Try this. http://www.cut-the-knot.org/pythagoras/cos36.shtml

**math**

plug in 3 for p: 3 = 14-2q

**Math**

distance = speed * time so just crank it out

**Math**

well, what numbers multiply to 40? example: 8*5 = 40 so, -8*5 = -40 8 * -5 = -40 Now, pick any other factors of 40. As long as an odd number of them are negative, the result is -40 2(-2)(-2)(-5) = -40 -10 * 4 = -40 and so on. Not as hard as it looked, was it?

**math**

both numbers are close to 2. What's 2+2 ?

**Maths**

well, he spent 1/4 + 1/3 = 7/12 so, what's left?

**College Algebra**

correct. Now it's smooth sailing.

**Precalculus**

so, you couldn't fix it and finish it off? The method was right in front of you ... r = 8sin? - 2cos? r^2 = 8r sin? - 2r cos? x^2+y^2 = 8y-2x That's ok as an equation, but I suspect you want a more standard form, so keep working it, completing the squares: x^2+2x + y^2...

**Precalculus**

x^2+y^2-6y-8 = 0 r^2 - 6rsin? - 8 = 0

**Precalculus**

I suspect a typo, but r = 8sin? - 2sin? r = 6sin? r sin? = 6sin^2? y = 6(x^2+y^2)

**math**

8/2 (2a+7d) = 50 50/2 (2a+49d) = 880 Solve for a and d, and then evaluate a+42d

**science**

decrease in distance, since F depends on 1/r^2

**math**

Since the terms of a GP have a common ratio r, (n+2)/(n-4) = (3n+1)/(n+2) If you solve for n, you will get two possible values. Use those to determine the appropriate values of r.

**College Algebra**

If you want to use elimination, arrange the equations so that x,y,z occur in order, and enter the coefficients into the array here: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

**College Algebra**

This one is easier to solve just using substitution: y=3z, so x+3z+z = 50 4.50x + 3(3z) + 4z = 4.38*50 Simplify those to get x + 4z = 50 4.50x + 13z = 219 Now, since x = 50-4z, 4.50(50-4z)+13z = 219 225-18z+13z = 219 5z = 6 z = 1.2 y=3z = 3.6 x=50-4z = 45.2 This makes sense, ...

**College Algebra**

you don't seem to be reading what is given. If you set up your x,y,z as you did, then x+y+z = 50 y = 3z 4.50x + 3y + 4z = 4.38*50

**College Algebra**

almost x + y = 5000 3x +7y = 23000

**College Algebra**

so, x+y=5000

**College Algebra**

No, 3x+7y = 2300 represents the total money collected. That equation is correct. How many tickets were sold?

**College Algebra**

Nope. x+y represents the total number of tickets sold. That is not 23000.

**College Algebra**

yes

**College Algebra**

You have tried to use the model where they give you the total interest earned. This one does not do that. Instead, if you read what was written, your second equation must be .08y = 32 + 2*.06x

**Maths**

hint: If 3|n^2 then 3|n*n

**Algebra**

Huh? -16*3^2+60*3+6 = 42

**Math**

the vertical mast makes a right angle with the level ground, right?

**Math**

50/AC = tan28° 50/AB = tan44° Now use the law of cosines to find BC

**Math**

sin^-1 (4x^4+x^2) = 1/6 pi take sine of both sides: 4x^4+x^2 = 1/2 8x^4+2x^1-1 = 0 (4x^2-1)(2x^2+1) = 0 ...

**Calculus**

If G(t) = ?g(t) dt then ?[3,x] t^5 dt = G(x)-f(3) so, f'(x) = x^5 ...

**algebra**

add up the amount of acid present: .40x + .10(540-x) = .20(540)

**math algebra three system equation**

If you want to see the details of elimination, a good place to go is http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

**Word problem**

if 16 of each, total is 480 Each quarter replaced with a nickel drops the total by 20. So, to lose 60 cents, replace 3 quarters with nickels. 19 nickels, 13 quarters 19*5+13*25 = 420 Too bad you won't show your work ...

**mechanies**

By the law of cosines, if the angle between a and b is ?, |a+b|^2 = |a|^2 + |b|^2 - 2|a||b|cos(180-?) So, |a+b|^2 = |a|^2 + |b|^2 + 2a•b

**Word problem**

or, say she has equal amounts of coins: 11 nickels, 11 dimes. The total amount is 11*5+11*10 = 165 Yet we are told that she has 175. Each time we replace a nickel with a dime, we get 5 more cents. So, to get those extra 10 cents, we need to replace 2 nickels with dimes: 9 ...

**MATH HELP PLS**

19. Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of his classes. He teaches band to 25 students. There are 48 students who take either band or choir or both. There are 9 students who take both math and choir with Mr. Green. Use ...

**MATH<<<<<**

I need help, how do you find the mean median mode and range?

**MATH<<<<<**

What are the mean, median, mode and range of the data set given the altitude of lakes in feet: –12, –9, –14, –39, –49, –49, –18, and –43?

**math**

49+51

**geometry**

the angle bisector splits a corner into two 45° angles. That means that the rectangle is 10x20.

**math**

note that ?BOC is a right triangle, since the diagonals are perpendicular.

**math**

so, did you do it? Start by labeling the time axis 1..2..3....10

**Math**

8gal/340mi * $4.00/gal = $0.0941/mi 7gal/350mi * $4.00/gal = $0.0800/mi

**Algebra**

You want to try that equation again? =y+76x2 makes no sense.

**Math**

#5 I have no idea, since I can't see the graph. It will slope up to the right, and pass through (0,3) and (-1,0) #6 (-5+10)/(-2+9) = 5/7

**Math**

permutations are just combinations where the order matters. combination: pick a 3-member committee from 10 candidates permutation: pick a president, vice-president, and secretary.

**Math**

2% growth means that x grows to x + .02x = x*1.02 The .02x is the amount of growth (addition) 1.02 is the growth factor (multiplication)

**Algebra**

what happens to each x and y to get from one to the other? The result is shifted 1 unit to the right and 2 units down So, the rule is (x,y) -> (x+1,y-2) and I have no idea of the meanings of "cornites" and "aigted"

**Math**

you know that on a rectangle, the top and bottom are equal, and that the two sides are equal, right? So that gives you 31-b = 2+a+2b 3a-b = 2+a+2b Make sure you have no typos, and then solve the system of equations.

**math**

The next x value could be 3, but how did you come up with y=7?

**math**

x=0, y=2*0^2+3 = 0+3 = 3 x=1, y=2*1^2+3 = 2+3 = 5 and so on for the other values.

**Maths**

if 1/4 are boys, the 18 girls make up 3/4 of the class. 18 = 3/4 * 24

**Mathematics**

The answers look good to me. Of course, as I have no idea what the questions were, that doesn't mean much...

**College Algebra**

(1) the graph of f(x-h) is shifted right by h. (2) The graph of s*f(x) is stretched by a factor of s. (3) reflection across the x-axis changes f(x) to -f(x). Now apply these rules to your mystery function.

**Math**

A(t)=600*(.5)^(t) this function gives a half-life of one year. As t grows by 1, the remaining amount is multiplied by 1/2. So, you want to adjust it so that the exponent grows by 1 as t changes by 5730. Take a look at the section in your text...

**Algebra**

what - you didn't like my earlier response? See the related questions below.

**Calculus**

you can find a by noting that tan(a) = -3/4 There will be a solution in QII and QIV

**math**

yeah - google can.

**Trigonometry**

well, in that case, as you have written it, there is no solution. There is no ? such that cos? = -306 If you can figure out what you really meant, learn about the 2nd function for the cos button on your calculator, and the arccos(?) or cos-1? function

**Trigonometry**

if you really mean 0612, then no. Otherwise, it is always possible to solve 2cos(x) = n if |n/2| <= 1

**Calculus II: Finding the volume of a revolution**

It is true that both sinx and cosx are bounded by -1 and 1, but the curves intersect at x = ?/4 So, the volume, using discs of thickness dx is v = ?[0,?/4] ?(R^2-r^2) dx where R=cosx and r=sinx v = ?[0,?/4] ?(cos^2x-sin^2x) dx = ?[0,?/4] ?cos(2x) dx = ?/2 Using shells of ...

**Calculus**

your reference angle is tan? = 1/?3 ? = ?/6 Now recall that tan? is positive in QI and QIII

**Calcalus**

Review your standard angles: 0, ?/6, ?/4, ?/3, ?/2 and signs in the quadrants Then solving these will be a cinch.

**Calculus**

cos? = x/r, so x = -7 r = 12 y = ?95 Now, recall that sin? = y/r tan? = y/x and let 'er rip. (a) and (c) should require no calculation at all...

**math**

The cup is a cone, not a cylinder. 9/12 = 3/4, so the volume is (3/4)^2 = 9/16 = 56.25% When 75% is full, the depth is ?.75 = 0.866, or 10.39 inches

**@Reiny saves the day**

nice save, Reiny. I later realized how I had misread the problem.

**geometry**

Huh? The angle bisector is the diagonal AC. The diagonals of a rhombus are perpendicular.

**Math**

The original dimensions are width: w length: 2w+3 (w+4)(2w+3+4) = 434+196 w=14 so, the new length 2w+3+4 = ?