# Posts by steve

Total # Posts: 51,759

**Math**

You have $13.67 dollar balance. You would divide that total by the .25 per minute. $13.67/.25 = 54.58 minutes or 54 minutes.

**Algebra2**

solve the equation cos theta - tan theta * cos theta = 0 for 0< theta < 2 pi My answer is pi/4,5pi/4

**Algebra2**

tan .09 = .09 so, the solutions are .09 + n? I suspect a typo. Did you mean tan^-1(-.09) ? and also on the last post?

**Algebra2**

Find the values of the inverse function in radians. tan^-1(0.09) answer choices: a.-0.09+2pi n b. no such angle exists c.-1.48+ pi n d. -0.09+ pi n My answer is D

**Algebra2**

thank you

**Algebra2**

Nope. Those answers are in QI and QIV The solutions are in QI and QII: sin(.71) = .65 so the solutions are .71 + 2n? pi-.71 + 2n? = 2.4 + 2n?

**Algebra2**

Find the values of the inverse function in radians. 1. sin^-1(0.65)? answer choices: a. 0.71+2pi n and -0.71+2pi n b. 0.71+2pi n and -3.85+2pi n c. 0.86+2pi n and -0.86 +2pi n d. -0.61+2pi n and 2.54+2pi n My answer is B

**Interest**

P * 0.034/12 * 13 = 50 solve for P

**VSU**

that would just be ?[0,30] 12 e^(0.2t) dt = 24,146

**Algebra II**

I assume you mean x-intercepts. Since e^0 = 1, you just need to solve x^2-9x+20 = 0 Back to Algebra I, now ...

**algbra**

add 4y to both sides add 14 to both sides divide by 4

**Algebra 2 help please.**

you don't want any maxima or minima, because that means the graph bends back on itself, meaning it fails the horizontal-line test. (A) is what you want. the domain is [0,pi]

**Calculus I**

Let t=0 at the given instant. The slope of the line is tan? (with ? measured clockwise from the x-axis) tan? = (7-3t)/(5+2t) = 7/5 sec^2? d?/dt = -29/(5+2t)^2 74/25 d?/dt = -29/25 d?/dt = -29/74 ------------------------------ second method ------------------------------ The ...

**Calculus Please Check my answer**

looks good to me.

**maths**

well, tan? = (24-1.5)/15 as seen from the man's eyes (assuming they are on the top of his head!)

**Calculus Please Check my answer**

looks good to me. Or, using shells, v = ?[0,1] 2?(y+1)(e^y-1) dy

**Math**

These are all easy, since the line segments are all horizontal or vertical. Just figure the lengths and compare. For example, AB = 2-(-4) = 6 They will be very easy to do if you just plot the points.

**Calculus**

Huh? There's nothing to do next. I gave you the first integral. The 2nd one was just for extra credit.

**Calculus**

The curve is maybe a bit easier to visualize if you write it as x = 4sin(y). The area is ?[0,?/2] (4-x) dy = ?[0,?/2] (4-4sin(y)) dy = 2?-4 ?4-4sin(y) dy = 4y + 4cos(y) This is a bit easier than using vertical strips, where the area is ?[0,4] arcsin(x/4) dx since integrating ...

**Maths**

Sorry. There's still a typo. Now you have two x terms. Don't you read what you post? As typed, your two circles do intersect. http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+-+10x+-+8y+%2B+18+%3D+0+,+x%5E2+%2B+y%5E2+-+8x+-+4y+%2B+14x+%3D+0 Rather than just ...

**Calculus**

well, f(x) is continuous, so the MVT applies. (f(1)-f(-1))/(1-(-1)) = (-8-8)/2 = -8 So, you want c where f'(c) = -8 3c^2-9 = -8 c = ±1/?3 Note that at f'(0) = -9, so it is too steep. You were just guessing, there, right? Just to show that our c values work, note...

**Math**

I don't know, but I'm sure google does ...

**Maths**

I suspect still a typo, since there are two x terms. Sorry, friend.

**Math**

5?700 = 5?(100*7) = 5(?100)(?7) = 5*10 ?7 = 50?7

**Maths**

are you ever going to fix the typo? Why keep wasting time with the same flawed posting? x^2 + y^2 - 8y - 4y + 14 = 0

**Math (written answer)**

T > -15 it doesn't get much simpler than this. Better review the section in your text... or google inequalities

**Calculus**

?3/4 (x^5/5 - x^4/2 + x^3/3) [0,1] = ?3/4 (1/5 - 1/2 + 1/3) = ?3/4 * 1/30 = ?3/120

**Calculus**

Not quite. You have figured the volume consisting of squares. For the triangle, the base is x-x^2, so the altitude is (x-x^2)?3/2. The volume is then ?[0,1] (1/2)(x-x^2)(x-x^2)?3/2 dx = ?[0,1] ?3/4 (x-x^2)^2 dx = ?3/120

**Algebra**

You know the sides are in the ratio 1:?3:2 Just switch angles to figure sin60

**MATH HELP PLZ**

what question? In any case, I'm sure it will mainly involve counting how many data fit the criteria, and then dividing by the total.

**Math**

2sinx + cosx = 0 2sinx = -cosx tanx = -1/2 Now you know there will be two solutions, in QII and QIV

**Algebra check plz**

Hmmm. What does they need to add a base shipping charge of $5 mean to you?

**math**

well, count up the ones with 4 C's and divide by the total number.

**Maths**

Or, try this way. The tangent line must be perpendicular to the radius which meets it. Also, the distance from the center to the line must be r. Since the center of the circle is at (a,b), if it touches the line at (h,k), we have m = (a-h)/(mh+c-b) That must have a single ...

**Maths**

If the line is tangent, the solution to the system of equations has a single solution. (x-a)^2 + (y-b)^2 = r^2 (x-a)^2 + (mx+c-b)^2 = r^2 x^2 - 2ax + a^2 + m^2x^2 + 2m(c-b)x + (c-b)^2 = r^2 (m^2+1)x^2 + (2mc-2mb-2a)x + a^2+(c-b)^2-r^2 = 0 To have a single solution, the ...

**Maths**

x^2 + y^2 - 10x - 8y + 18 = 0 x^2-10x + y^2-8y = -18 x^2-10x+25 + y^2-8y+16 = -18+25+16 (x-5)^2 + (y-4)^2 = 23 The second equation has a typo. Fix it and complete the squares to find its center. Then compare the distance between centers to the sum of the radii.

**Maths**

(a) 4/9 * 3/4 = 1/3 (b) 1 - 1/4 - 1/3 = 5/12

**math**

x = 2/10 x + 1/2 x + 12 6/20 x = 12 x = 40

**math**

since the scale ratio remains constant, x/135 = 1/20

**math**

(10/12 - 2/3) / (1/6) = 2 days

**math series2(please helpassessment tomorrow)**

If M = bL^3/12 replace b by 1.015b and L by 0.98L and you have (1.015b)(0.98L)^3/12 = (1.015*0.98^3)bL^3/12 = 0.955M so, M has decreased by 4.5% I see no reason to resort to binomial approximation. But, if you want to try it, and compare results, feel free.

**math**

in 10 days, A does 10/12 of the work In 2 days, B destroys 2/3 of the work 10/12 - 8/12 = 1/6 Since A does 1/12 each day, it will take 2 more days to finish.

**geography**

well, since there are 360 degrees in a complete circle, and there are 24 hours in a day, each hour of time shift involves 15 degrees of longitude.

**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