# Posts by Steve

Total # Posts: 51,716

**math**

what do you mean by "strength" of class?

**Science**

.0050 * 53.2 = ?

**math**

geez - impatient much?

**maths**

where ever did you get those numbers? 4 pi r^2 dr/dt = 4 * pi * 5^2 * 4 = 400 pi cm^3/s

**maths**

v = 4/3 pi r^3 dv/dt = 4 pi r^2 dr/dt now just plug in your numbers ...

**Algebra**

say that all 21 are chickens. That would give 42 legs. Each chicken that is replaced with a cow will raise the leg count by 2. we have 20 extra legs, so that means we need 10 cows. Thus, 11 chickens, 10 cows

**Algebra2**

A sound wave is modeled with the equation y = 1/4 cos 2pi/3 ?. a. Find the period. Explain your method. b. Find the amplitude. Explain your method.

**Algebra2**

A wave is modeled with the function y = 1/2 sin 3?, where ? is in radius. Describe the graph of this function, including its period, amplitude, and points of intersection with the x-axis. My answers are Amplitude = 1/2 Period = 2pi/3 Intersection= (0,0), (?/3,0) and (2?/3,0)

**MATH**

see related questions below.

**Algebra2**

thank you

**Algebra2**

3tan(?/3) = 8 tan(?/3) = 8/3 Now, tan(1.212) = 8/3, so use that as your reference angle. tan? is positive is QI and QIII, so ?/3 = 1.212 or ?+1.212=4.353 so, ?=3.636 or 13.061 The only solution in [0,2?) is 3.636

**Algebra2**

3tan 1/3 theta=8 in the interval from 0 to 2pi. Solve the equation.

**Algebra2**

I think it is 172 ft^2

**Algebra2**

Since the height is 5tan?, the side length must be 10. Draw a diagram to see this. Now you have the base and height of each triangle, so crank it out.

**Algebra2**

A pentagon can be divided into 5 congruent triangles. The function y=5 tan theta models the height of each triangle. What is the area of the pentagon if theta = 54 degrees?

**maths - ouch!**

That pesky zero key! Grrr

**maths**

300/1500 = 1/5 = 2%

**Algebra2**

Thank you

**Algebra2**

Can't tell just what you mean. Ever heard of notation for powers (^) of arguments () ? ? = ?/4 sin(2?) = sin(2*?/4) = sin(?/2) = 1 C is correct if you meant sin^2(?)

**Algebra2**

Use the graph of y=sin2theta to find the value of sin 2 theta for theta=pi/4. a--1 b-0 c-0.5 d-1 my answer is -1

**Algebra**

well, 2? radians every minute, so how many minutes have passed?

**MAth**

(3a+4)/(4a+4) * (a^2-6a-7) = (3a+4)/(4(a+1)) * (a+1)(a-7) The (a+1) factors out of top and bottom, leaving = (3a+4)(a-7)/4 = (3a^2-17a-28)/4

**Physics**

if the "x" axis is for time, then a negative position means that motion has backed up past where it started.

**Maths**

If he started with x, then x - x/3 - 12 = x/2

**algebra**

(4/?50)/?(189x) * ?(189x)/?(189x) = (4?50 ?189x)/(189x) = 4?(50*189x)/(189x) = 4?(25*2*9*21x)/(189x) = 4*5*3?(2*21x)/(189x) = (20?(42x))/63x Somehow I suspect a typo

**math**

divide one by the other, I guess. Pretty vague question

**Math (written answer)**

c/4 - 5 = 4 add 5 to get all the c stuff on one side: c/4 = 9 now multiply by 4 to clear the fraction: c = 4*9 c = 36

**discrete math**

(a) you have 5 1s and 3 0s so, use the formula for permutations of a set with duplicates: 8!/(5!3!) = 56 That's because for every one of the 8! permutations, the 1s and 0s can be shuffled without altering the string. (b) and (c) are similar.

**discrete math**

Assuming that k from 1 to n fails to execute if n < 1, when i=1, j fails when i=2, k fails at j=1 when i=3, j=1..2, k=0+1..1=1 when i=4, j=1..3, k=0+1+2=3 when i=5, j=1..4, k=0+1+2+3=6 ... when i=n, j=1..n-1, k=0+1+2+...+n-2 = (n-2)(n-1)/2 so, the total prints executed is 1...

**Calculus**

I assume you have drawn the region. It's a triangular shape with vertices at (0,1), (1,1) and (1,e). So, using discs (washers) of thickness dx, v = ?[0,1] ?(R^2-r^2) dx where R=y+2 and r=3 v = ?[0,1] ?((e^x+2)^2-3^2) dx = ?/2 (e^2+8e-19) Or, using shells of thickness dy, v...

**Math**

2.5/1.5 = ?

**math**

time = distance/speed just use your numbers.

**MATHS**

In triangle BCF, sides BC and BF are equal, since they are both the length s of one side of the pentagon. angle B is 18° (90°-72°, the exterior angle of the pentagon) So, since BCF is isosceles, angle BCF = (180-18)/2 = 81°

**math**

3/4 of 24 = 18 dozen so , ...

**Algebra 1**

that's better

**math**

so, pick one.

**Calculus**

the distance is s = ?[0,2] -2t^2 + 4 dt = -2/3 t^3 + 4t [0,2] = -16/3 + 8 = 8/3 you are correct

**Calculus**

The curves intersect at x = ±0.824 Since the area is symmetric we can just work with one half of it, and double that value: a = 2?[0,0.824] cosx - x^2 dx

**Math**

which pair of factors of 96 add up to 22? (p+6)(p+16) which pair of factors of 63 differ by 18? (q+21)(q-3)

**GEOMETRY**

It's certainly possible, but I'm not going to work through all the possible combinations of sides to come up with 650. Why not just show us your work, describe the sides of each figure, and we can check the accuracy.

**Maths**

1 2 3 0.1 2 2.1 1.9 2 3.9 Just pick any two numbers that differ by 2; one above 2 and one below.

**Math algebra**

a+15 = 4a a=5 check: in 15 years she will be 20=4*5

**Calculus**

The graphs intersect at (0,2) and (5,2). So, the area is just the region below the parabola and above the line: a = ?[0,5] y-2 dx = ?[0,5] (-x^2+5x+2-2) dx = 125/6

**discrete math**

(a) 36^5 (b) 10*36^4 (c) 10*9*9*9 (d) assuming you just mean that the exact sequence of 3 letters does not repeat, then 36P3 * (36P3 - 1) extra credit: what if you mean that the first 3 letters do not all repeat in the 2nd half of the password?

**discrete math**

huh? I read the problem. It says two circuits are considered the same if they have the same output bit for every possible input string There are only two possible output values for a single bit, so ...

**discrete math**

Well, since there are only two outputs possible, I'd say that there are only 2 "different" circuits, regardless of the number of bits read in.

**Math**

what is the shape of the truss?

**maths!**

Sorry - I've kinda lost track of what you're referring to. I don't recall using any foreign alphabets. Maybe you could refresh my memory here ...

**math**

I guess that would depend on the size of the box, dontcha think?

**maths**

Start at -20 and move right to +5 How far have you moved? That's how many years had passed

**Math-Trigonometry**

A Ferris wheel has a radius of 37.8 feet. so, the amplitude is 37.8, and you can start with f(t) = 37.8 sin(kt) The bottom of the Ferris wheel sits 0.7 feet above the ground. So, the axle is 37.8+0.7 = 38.5 feet off the ground: f(t) = 38.5+37.8 sin(kt) You board the Ferris ...

**Calculas**

approximate how? There are lots of methods. c-e = 2.74-e = 0.00217 (c-e)/e = 0.00798 ? 0.8%

**math-precalculus**

A Ferris wheel is 50 meters in diameter amplitude is 25, so y = 25sin(kt)+k The boarding platform is 1 meter from the ground, so the axle is 26 feet up y = 26+25sin(kt) The period is 6 minutes, so 2?/k = 6 ==> k = ?/3 y = 26+25sin(?/3 t) If we assume that at t=0 we are ...

**Maths**

f-15 = 3(s-15) f-19 = 4(s-19) solve for s and f, and then find x such that f-x = 2(s-x)

**algebra**

V = k/P, so PV = k you want P such that 180P = 250*35 By the way, pressure is in Pascals = N/m^2, not kg/cm^3 kg/cm^3 is density

**Probablity**

Since there is replacement, all the events are independent. So, for n draws, P(any specific n colors) = (1/7)^n

**math**

just do the dot product. You get cosx*cos(a+x) + sinx*sin(a+x) = cosx*cosa*cosx - cosx*sina*sinx + sinx*sina*cosx + sinx*cosa*sinx = cosa(cos^2x+sin^2x) = cosa

**Math**

Draw the figure. Since AP and BQ are parallel, PABQ is a trapezoid. Draw PR to intersect BQ at R. AP=BR=4 so RQ=5 So, the sides are PA=4 AB=PR=13 BQ=9 QP = ?(5^2+13^2) = ?194

**math URGENT HELPPPP!**

With a slant height of 12, each triangular face has a base of s and a height of 12, so 3(12s/2) = 198 s = 11

**math**

no. try it. it's easy to demonstrate.

**Math**

well, 363 is 2 std below the mean. So, apply your rule.

**discrete math**

there are 2^n possibilities for n bits.

**mathvariation 2(steve,reiny,reed) pls I need help!**

E = a+bL 1.4 = a+0.02b 2 = a+0.03b subtract, and you have 0.6 = 0.01b b = 60 so a = 0.2 E = 0.2 + 60L

**variation! Steve pls check my solution!!!**

there seems to be a typo 3 lines up from the end... But the math looks good, sorta.

**GGHSS COTTON HILL Maths**

clearly, d = -6 and, a=57 Now you can take it from there ...

**maths**

Either the 1000 should be 100 or the answer is -951

**math**

Just start trying factors, up to the square root of the number. 3|953? no clearly 5 is not a divisor 7|953? no trying all the primes up to 30, we see that there are no factors of 953. It is prime. We stop at the square root, because if there is a factor greater than ?n, then ...

**Math**

sure, but not likely. If the scores were something like 100,100,100,100,100,100,100,100,5,5 sum=810, so mean=81

**Math**

the length and width must add up to 8, so the only possible dimensions are 1x7 2x6 3x5 4x4

**Calculus**

4cos^2(t)sin^2(t) + sin^2(t) = 4(1-sin^2(t))sin^2(t)+sin^2(t) = 4sin^2(t)-4sin^2(t)+sin^2(t) = 5sin^2(t)-4sin^4(t) = sin^2(t) (5-4sin^2(t)) = sin^2(t) (1+4cos^2(t)) So, now you have ??(1+4cos^2(t)) sin(t) dt u = cos(t) du = -sin(t) dt -??(1+4u^2) du Now let 2u = tan? and see ...

**Calc**

r = 5 csc? r sin? = 5 y = 5

**Math**

Dunno about tape diagrams, but if the 1st pile has x coins, then x + 3x + 3x+50 = 680 x = 90

**Calculus**

see whether this helps: http://www.wolframalpha.com/input/?i=sum(k%3D1..n)+(k%5E2%2B1)%2Fk%5E4 You can plug in 5 or 9 for n and get the partial sums...

**@bobpursley gets it right**

dang - I meant the angular speed is the rotational speed (rev/s) times 2?.

**physics help**

each revolution travels 2?r = 68? cm. So, 3900cm/s * 1rev/68?cm = 3900/68? rev/s That gives you frequency. Flip it upside down to get the period. the angular speed is the linear speed divided by 2?.

**Math**

google is your friend. Try searching for circle area circumference and so on, and you will get many explanations and examples.

**Calculus**

using shells of thickness dx, we have v = ?[1,2] 2?rh dx where r=x and h=y-1 v = ?[1,2] 2?x(x^3-1) dx = 47?/5 using discs (washers) of thickness dy, v = ?[1,8] ?(R^2-r^2) dy where R=2 and r=x v = ?[1,8] ?(4-y^(2/3)) dy = 47?/5

**math**

each face has a base of 30/4 = 7.5 So, there are 4 edges of (27.5-7.5)/2 = 10

**math**

I think it was sin130.tan60 ÷(cos540.tan230.sin400) = (sin50 ?3)/(-1 * sin50/cos50 * sin40) = -?3 cos50/sin40 But since cos50 = sin40, the result is just -?3

**Math**

all those nice right angles and 45-degree angles make things easy. If you draw the figure, you can easily see that 1/2 the diagonal is 2+3 = 5

**geometry**

well any square's diagonal is ?2 times its side.

**Math**

all pyramids have triangular faces and polygonal bases.

**Math**

another prism. All prisms have two bases and some number of (usually) rectangular lateral faces.

**math**

2800 rev/min * 24?cm/rev = 67200? cm/min I'm sure you can convert that to m/s

**Math**

1000*0.60 + 2000*0.80 = ? I'm guessing at the 80c notebook price, but you can use whatever you think it will be.

**Math**

this site should help: http://davidmlane.com/hyperstat/z_table.html

**Math**

prism with 7-sided bases

**math**

using only angles, congruence is not possible. The best you can do is similarity.

**Calculus**

dT/dt = -kT-A dT/(kT+A) = -k dt 1/k ln(kT+A) = -kt + C T(0) = 90 so 1/k ln(90k+30) = C T(1) = 85 so 1/k ln(85k+30) = C-k Solve for C and k, then evaluate t when 1/k ln(60k+30) = C-kt

**Calculus**

dy/dx = x/y y dy = x dx y^2 = x^2 + c using the point (2,1) 1 = 4+c c = -3 x^2-y^2 = 3

**Math Correction**

I forgot that the perpendicular bisector passes through the midpoint of PQ: (3,-2). So, its equation is y+2 = 1/3 (x-3) So, the center of the circle is at (0,-3). Now find the radius, and you're about done.

**sir steve reiny damon bob maths help steve!!!**

(a) the perpendicular bisector of the chord PQ contains the center of the circle. Since PQ has slope -3, the perpendicular will have slope 1/3. So, the line is y-1 = 1/3 (x-2) Similarly, the radius through Q is perpendicular to the line 2x-y=13. Its equation is y+5 = -1/2 (x-4...

**math**

recall that the vertex of a parabola is at x = -b/2a = 480/0.12 = ?

**math**

8+y <= -22

**Maths**

well, .86666666 - .860000 = .0066666 = 2/300 hours so route 2 is faster by 2/300 hours, or 24 seconds

**Maths**

time = distance/speed So, do the divisions and compare the times.

**Mathematics.**

clearly, the whole number part must be as small as possible, so it'd be 2 ?/7 x 3 1/? Now we want as large a denominator as possible, so I'd go with 2 4/7 x 3 1/5 = 8 8/35 Hmmm/ Nope, 3 4/7 x 2 1/5 = 7 6/7

**Math**

I think B is more likely. All the other pairs have two numbers approximately equal. 4.2 is over twice as big as 2.

**science**

draw a diagram. This is nice and easy, since you have a right triangle. The distance traveled by the plane in t hours must form the hypotenuse of the triangle, so (40t)^2 = (20t)^2 + 500^2 t = 25/?3 = 14.434 hours. Now you can figure the distance and the angle ? (west of north...

**Math**

all regular pentagons are similar.