Wednesday

July 27, 2016
Total # Posts: 1,781

**chemistry**

Your post in incomplete. What is your question?
*January 28, 2011*

**math**

Assuming you earn a total of $70. x = amt earned Sat 2x = amt earned Sun 1/2 x = amt earned Fri x + 2x + 1/2 x = 70 Solve for x
*January 28, 2011*

**Income Tax**

My answer above is assuming you meant the United States. I don't know, if your question is for a different country.
*January 28, 2011*

**Income Tax**

I would say true.
*January 28, 2011*

**Algebra**

MP = (1,3), A = (8,2), B = (x2,y2) Midpoint x = 1/2(x1 + x2) Midpoint y = 1/2(y1 + y2) 1 = 1/2(8 + x2) 2 = 8 + x2 x2 = -6 3 = 1/2(-2 + y2) 6 = -2 + y2 y2 = 8 B = (x2,y2) = (-6,8)
*January 28, 2011*

**geometry**

By "around" do you mean perimeter? A six figured polygon with equal sides, is called a regular hexagon.
*January 28, 2011*

**economics**

then? Is this problem complete?
*January 28, 2011*

**math**

Not subtraction. Division would be used to solve this problem.
*January 28, 2011*

**math**

Convert all the fractions to fractions with the same denominator, using the LCD. A. 18/30 B. 20/30 C. 15/30 D. 21/30 Now, wouldn't the largest fraction be closest in value to 1?
*January 28, 2011*

**math**

Is this the exact wording of the problem? Assuming "then subtract from result" is referring to the unknown number n, 7(n/3) - n = 36 7n - 3n = 108 4n = 108 n = 27
*January 28, 2011*

**Business Math**

I = PRT (Interest = Principal * Rate * Time) .04 interest/yr = .003 mo (.04/12) I = 6000 * 0.003 * 7 You can take it from here.
*January 28, 2011*

**Integrated Physics and Chemistry**

I found Ideal Mechanical Advantage.
*January 28, 2011*

**geometry **

Your post was answered below. Please don't repost a question without first checking to see if your 1st post was answered.
*January 28, 2011*

**Mathl**

Please put the school subject name in the "School Subject" box. Thank you.
*January 28, 2011*

**geometry**

Thanks drwls. I could not understand this post at all. :)
*January 28, 2011*

**geometry**

See your first post.
*January 28, 2011*

**bedmass**

There is no need to multiply the left side of the equation by 2.
*January 28, 2011*

**Math**

If you would like a math tutor to help with your problem in a timely manner, please follow directions and put the school subject in the "School Subject" box.
*January 28, 2011*

**geometry**

I do not understand your post. Exterior angle? Where? Which one? "find the other interior opposite angle" Other? Interior? Opposite what? "and the third angle of each triangle" How many angles do you expect someone to find? Please re-read your post. Think ...
*January 28, 2011*

**Algebra B**

You need to use parentheses. x-4/x+3 = x-1/x+2 Is the problem, x - 4/(x+3) = (x-1)/x + 2 x - 4/x + 3 = x - 1/(x+2) Or x - 4/x + 3 = x - 1/x + 2 Can you see the difference in these equations? This is probably why no one answered your question.
*January 27, 2011*

**math 9 helppp**

Well, you arrived at the same answer, but the equation is not correct for the problem. Your equation doesn't make sense. Where did you get 380? 380/7 = 54.29 not 55, even rounded it would be 54 or 54.30 not 55. And, it is 3.50 fee/ticket. You can't randomly subtract 3....
*January 27, 2011*

**math 9 helppp**

The equation is 7x = 385 - 7(3.50) 7x = 385 - 24.50 7x = 360.50 x = 51.5 EACH ticket has a 3.50 service fee
*January 27, 2011*

**Calc.**

| x^3*sqrt(x^2 + 1) dx | = integral sign | x^2 (sqrt(x^2 + 1)) x dx u = x^2 du = 2x dx 1/2 du = x dx 1/2 | u (sqrt(u + 1)) du w = u + 1 dw = du u = w - 1 1/2 | (w - 1) (sqrt(w)) dw 1/2 | (w - 1) w^1/2 dw 1/2 | w^3/2 dw - 1/2 | w^1/2 dw 1/2 (2/5 w^5/2) - 1/2 ( 2/3 w^3/2) + C 2/...
*January 27, 2011*

**algebra**

Not correct 121t^4 + 44t + 4
*January 27, 2011*

**Math**

I pasted below an example that reiny (tutor here) did recently. 10 and 55 are done in this example. Follow what reiny did for 45. EXAMPLE 10 = 2*5 75 = 3*5*5 55 = 5*11 The GCF is the largest number that divides into all three numbers, in this case 5 the GCF ≤ the ...
*January 27, 2011*

**math**

(6x^2-3x-9)-(2x^2+8x+3) Remove parentheses and change the sign of the terms inside the parenthese Combine like terms 6x^2 - 3x - 9 - 2x^2 -8x - 3 I leave the rest to you
*January 27, 2011*

**algebra**

Add the two 3x^2 + 7x - 3 + 5x + 2 = ?
*January 27, 2011*

**Geometry**

Correct
*January 27, 2011*

**algebra**

2z^2 IS 2z^2 or 2*z*z
*January 27, 2011*

**Math**

The time x cannot be negative. You must have made a mistake. The two equations are, distance gone by car = 64x distance gone by truck = 56(x+1/4) Since the distances are equal, 64x = 56(x + 1/4) 64x = 56x + 56/4 64x = 56x + 14 8x = 14 x = 1.75 From above (Reiny's ...
*January 27, 2011*

**math**

If P1(x1,y1) and P2 (x2,y2) are two points on a line, the number m defined by the equation Slope m = (y2 - y1)/(x2 - x1) is called the slope of the line.
*January 27, 2011*

**math**

3.4 + (5x - 1.7) plug 2.2 in for x 3.4 + 5(2.2) - 1.7 3.4 + 11.0 - 1.7 You can take it from here
*January 27, 2011*

**MATH**

Slope-intercept form y = mx + b m = slope b = y-intercept 5X - 2Y = -10 2y = 5x + 10 y = 5/2 x + 5 Slope m = 5/3 y-intercept b = 5
*January 27, 2011*

**Pre-Algebra Math**

$1167 ttl value of bills $147 ttl value of other bills $1020 value of tens and twenties n = number of tens and twenties 10n = ttl value of tens 20n = ttl value of twenties 10n + 20n = 1020 Solve for n, number of tens and twenties (equal number of both)
*January 27, 2011*

**Algebra**

2x + 5y = 4 Add -2x to both sides 5y = -2x + 4 divide both sides by 5 y = -2/5 y + 4/5 So, slope m = -2/5 Perpendicular lines, the slopes are negative reciprocals of each other. Slope = -2/5 m1 * m2 = -1 -2/5 * m2 = -1 m2 = -1/(-2/5) m2 = 5/2 Now you have to find b, for the ...
*January 27, 2011*

**Math**

I'm not sure what your question is ? 1 mile = 5280 feet 2 miles = 2 * 5280 ft = ? feet You can do the multiplication
*January 27, 2011*

**math**

Perimeter ABCD = 54 L = 12 in Perimeter LMNO = x L = 2.4 ft = 28.8 in P = 2w + 2L Use Ratio to find x (perimeter LMNO) 12/28.8 = 54/x Cross multiply 12x = 1555.2 x = 129.6 LMNO P = 129.6 L = 28.8 P = 2w + 2L 129.6 = 2w + 2(28.8) Solve for w, width of LMNO
*January 27, 2011*

**Calc.**

After all that, I confused myself and did the wrong problem I think. IF your problem is |x cos(x^2) dx just like your 2nd reply says, ignore the above. I wrote down | x cos (x^2) but integrated | x cos^2 x dx So, your very first post IS correct if it is cos x^2 and not cos^2 x...
*January 27, 2011*

**Calc.**

| x cos(x^2) dx The first line should be | x cos^2 x I forgot to change it.
*January 27, 2011*

**Calc.**

| x cos(x^2) dx | = integral symbol By half-angle cos^2 x = 1/2 (1 + cos 2x) | x 1/2 (1 + cos 2x) dx 1/2 | x + x cos 2x dx 1/2 | x dx + 1/2 | x cos 2x dx 1/4 x^2 + 1/2 | x cos 2x dx Integrate | x cos 2x dx by Integration by Parts u = x, dv = cos 2x dx du = dx, v = 1/2 sin 2x 1...
*January 27, 2011*

**Calc.**

S b= sqrt(Pi) a= 0 xcos(x^2)dx What is "S b = sqrt(Pi) a= 0" What are you integrating? Is it this? x cos(x^2) dx ? from a = 0 to b = (sqrt(pi))?
*January 27, 2011*

**Bobpursely**

Thank you, drwls!
*January 27, 2011*

**Bobpursely**

FYI-Many questions are under the name "Anonymous". If you expect a particular tutor to help you, you need to be more specific.
*January 27, 2011*

**Help**

sorry about the typo!
*January 27, 2011*

**Help**

Yes, x = 3 See your 1st post. I had a typo, and corrected it below.
*January 27, 2011*

**Help**

This problem was solved below for you. How many tutors do you need to solve this for you?
*January 27, 2011*

**college algebra**

1500sqrt(2)+1450 unless you use the value of (sqrt(2)), these two expressions cannot be added.
*January 27, 2011*

**Math**

Ignore above. I thought it was 3x = 1, not 3x - 8 = 1 x = 3
*January 27, 2011*

**Math**

Correct!
*January 27, 2011*

**Math**

(cubed root 3x-8)-1=0 Take the cube rt of both sides 3x - 1 = 0 3x = ? I'll leave this for you to finish
*January 27, 2011*

**Math correct or not**

Not correct
*January 27, 2011*

**Math**

Are you asking for the area of this box? A = Lw A = 2 2/3 * 2 2/3 A = 8/3 * 8/3 A = 64/9 = 7 1/9 sq ft
*January 27, 2011*

**algebra**

s = 4x - 3 V = s^3 V = (4x - 3)^3 V = (4x - 3)^2 (4x - 3) I leave the multiplication to you.
*January 27, 2011*

**finance**

This question asks "from your personal experiences". Post your personal experiences and a tutor will offer comments.
*January 27, 2011*

**7th grade math: Pythagorean Thereom**

It could be, 9 + 40 = 49 a^2 + b^2 = 41^2 a + b = 49 Solving simultaneously a = 9, b = 40 or a = 40, b = 9 41^2 = 9^2 + 40^2 1681 = 81 + 1600 1681 = 1681
*January 27, 2011*

**7th grade math: Pythagorean Thereom**

Do you mean that the two unknown sides total 49, or all three sides total 49?
*January 27, 2011*

**Gym**

Going up, down, diagonal what? Adding, multiplying, etc to equal 24? What exactly does the problem state to do?
*January 27, 2011*

**trig**

Correct!
*January 27, 2011*

**Precalc**

No solutions exist
*January 27, 2011*

**Algebra 1**

2x + 5y = 4 Add -2x to both sides 5y = -2x + 4 divide both sides by 5 y = -2/5 y + 4/5 So, slope m = -2/5 Perpendicular lines, the slopes are negative reciprocals of each other. Slope = -2/5 m1 * m2 = -1 -2/5 * m2 = -1 m2 = -1/(-2/5) m2 = 5/2 Now you have to find b, for the ...
*January 27, 2011*

**algebra**

R = 221.19x - 0.2x^2 R = 221.19 (273) - 0.2(273)^2 R = 221.19 (273) - 0.2(74,529) R = ?
*January 27, 2011*

**Math **

To understand this you really need to sketch this out. Think of the x/y axis on a graph, with the y-axis as North and the x-axis as East. Both boats are traveling from the North (top of the y-axis), toward East (down toward the x-axis). The first bearing, forms a 47 deg. angle...
*January 27, 2011*

**Exponents**

Meant to comment, If you square x^-2, x^-2 * x^-2 = x-4 = 1/x^4 1/^4 = 81 square root of both sides 1/x^2 = 9 You are back to the above.
*January 27, 2011*

**Exponents**

x^(-2) = 9 Since, x^(-2) = 1/x^2 1/x^2 = 9 Multiply both sides by x^2 1 = 9x^2 1/9 = x^2 square root of both sides + - 1/3 = x
*January 27, 2011*

**Math EASY**

Like Ms. Sue said. I just did this based on info you provided.
*January 27, 2011*

**Math EASY**

(83 + 3/5 + 5/5)/3 (83 + 60 + 100) = 243/3 = 81 81%
*January 27, 2011*

**adult education-Psycology?**

If you would like help in a timely manner please follow directions and put your school subject in the "School Subject" box.
*January 27, 2011*

**MATH**

If I am picturing this correctly, a right triangle is formed, with the hypotenuse c = 75 cm. Since the top of the wall is 5.4 m above ground, 5.4 - side a = height above ground to the roof edge. Angle A = 180 - (23 + 90) = 82 degrees sin 82 = op/hyp = a/c = a/75 0.9903 = a/75 ...
*January 27, 2011*

**arithmetic**

5/6 * 4 Since 4 = 4/1, you multiply the numerator of 5/6 times 4, and the denominator of 5/6 * 1 5/6 * 4/1 = 20/6 = 10/3
*January 27, 2011*

**5th math**

x = number of boys x + 4 = number of girls x + x + 4 = 30 2x + 4 = 30 2x = 26 x = 13 x = number of boys = 13 x + 4 = number of girls = 13 + 4 = 17 13 boys + 17 girls = 30
*January 27, 2011*

**5th grade**

Pat, I read that wrong. 77/35 * 285 = ?
*January 26, 2011*

**5th grade**

All that info and all the problem wants is the cost for 285? .77 each x 285 = ?
*January 26, 2011*

**5th math**

w = width, L = length Perimeter = 2w + 2L 60 = 2w + 2L 60 = 2(w + L) Divide both sides by 2 30 = w + L Since a square has equal sides and w + L = 30 15 + 15 = 30 So each side is 15 15 + 15 + 15 + 15 = 60 I don't know if a 5th grader does this type of problem like I did. If...
*January 26, 2011*

**Algebra 1**

2x + 5y = 4 Add -2x to both sides 5y = -2x + 4 divide both sides by 5 y = -2/5 y + 4/5 So, slope m = -2/5 Perpendicular lines, the slopes are negative reciprocals of each other. Slope = -2/5 m1 * m2 = -1 -2/5 * m2 = -1 m2 = -1/(-2/5) m2 = 5/2 Now you have to find b, for the ...
*January 26, 2011*

**Algebra 1**

Did you see the answer to your post below? Post your work so I can see where you are having trouble.
*January 26, 2011*

**5th math**

Mean is the average ADD the number of letters and divide by the number of numbers. 105 + 97 + 88 + 86 = sum Average = sum/4
*January 26, 2011*

**5th math **

28.5 min for 4 miles 28.5 min/4 mi = 7.125 min each mile So, this is your equation 4k = 28.5 k = 28.5/4
*January 26, 2011*

**algebra**

-4x - 12 = 0 Add 12 to both sides -4x - 12 + 12 = 0 + 12 -4x = 12 Divide both sides by -4 x = ?
*January 26, 2011*

**Algebra 1**

x + 6 = length of sidewalk x + 4 = width of sidewalk A = Lw A = (x + 6)(x + 4) (x + 6)(x + 4) = 39 Expand and solve for x
*January 26, 2011*

**pre algebra-math**

helper - please choose another 'Name' I've been using this name and it is very confusing when you use the same name. Thank you so much. Just put a number next to the name, if you want to use 'helper'.
*January 26, 2011*

**pre algebra-math**

yes^ please help 18/12 and 3/27
*January 26, 2011*

**...**

you mean 18/12 then the other question is 3/27!!!
*January 26, 2011*

**Algebra 1**

(6,7);6x+y=9. y = mx + b 6x + y = 9 y = -6x + 9 m = -6 m1*m2 = -1 -6*m2 = -1 m2 = 1/6 So, the slope of the perpendicular line is 1/6. y = 1/6 x + b Now find b with point (6,7). y = 1/6 x + b 7 = 1/6 (6) + b 7 = 6/6 + b 7 = 1 + b b = 6 y = 1/6 x + b y = 1/6 x + 6
*January 26, 2011*

**Algebra 1**

Your equation is not complete. 6x + = 9?
*January 26, 2011*

**ALGEBRA**

V = e^3 (e = edge) 64 = e^3 e = 4 Can you take it from here?
*January 26, 2011*

**Algebra**

Remaining angle = 180 - (55 + 47) Supplementary angles total 180 degrees You should be able to try this now.
*January 26, 2011*

** 7th grade math**

Area = ((b1 + b2)/2)* h 4 = ((b1 + b2)/2)* 1 4 = ((b1 + b2)/2) Multiply both sides by 2 8 = b1 + b2 So, the sum of base1 and base2 = 8 I don't know what you mean by 'best numbers'?
*January 26, 2011*

**Algebra**

Find the mean (average) 1 + 2 + 4 + 5 + 6 + 8 + 9 + 10 + 10 = sum Average = sum/9 The median is the middle value, 6 mean - median = postive difference Then express this difference as a fraction
*January 26, 2011*

**Pre Calc**

If I am understanding this picture, a right triangle is formed with angle A = 50 degrees and side b = 23 (15 + 8). You need the length of a, which is the streetlamp. tan A = opp/adj = a/b = a/23 tan 50 = a/23 1.19 = a/23 Solve for a, for your answer
*January 26, 2011*

**math**

x = one angle 2x = largest angle 1/3 x = smallest angle x + 2x + 1/3 x = 180 Solve for x
*January 26, 2011*

**Reading APE**

Did you read this? Tell us what you think and someone will proof your answer.
*January 26, 2011*

**algbra**

You need more parentheses in your problem. t = (r + k)/(2w)? t = r + (k/(2w)) ? t = r + (k/2) w ? Or?
*January 26, 2011*

**Algebra**

Answer is 2a^(-4)c^4 a^6/a^10 = a^(6-10) = a^-4
*January 26, 2011*

**Precalc**

You're welcome!
*January 26, 2011*

**Precalc**

solve 5(2^(3x)) = 8 5(2^(3x)) = 8 2^3x = 8/5 2^3x = 1.6 take logarithms of both sides 3x log 2 = log 1.6 Divide both sides by 3 log 2 x = log 1.6/(3 log 2)
*January 26, 2011*

**Pre-Algebra**

2/3 m - 5 = n Multiply both sides by 3 2m - 15 = 3n 2m = 3n + 15 m = ?
*January 26, 2011*

**L Pre-Cal**

3x-5y+5=1 5x-2y+3z=0 7x-y+3z=0 Do you have a typo? 3x - 5y + 5? = 1 There is no z term in equation 1?
*January 26, 2011*

**pythagrean theorem**

Ms. Sue is correct. I get 12.5 too.
*January 26, 2011*

**math**

Convert a) through e) to addition expressions Change the sign to + and change the sign of the number to the opposite of what it is. a) -6 - 9 = -6 + (-9) b) -10 - (-12) = -10 + 12 Now you do c), d) and e)
*January 26, 2011*

**math**

m = (-1-3)/(2-(-4) You had 3, should be 2
*January 26, 2011*