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December 20, 2014

December 20, 2014

Total # Posts: 1,701

**alg 2**

x^3/4 * x^3/4 * x^3/4 = x^9/64
*January 20, 2011*

**5th grade math**

w = 78 L = 68 p = 2w + 2L p = 2(78) + 2(68) can you do this now?
*January 19, 2011*

**algebra 2**

v^2 + 8v + 6 = 0 this cannot be factored either complete the square or use the quadratic formula to solve
*January 19, 2011*

**algebra 2**

quadratic formula x = (-b +- (sqrt(b^2 - 4ac))/2a 5x^2 + 3 = -7x 5x^2 + 7x + 3 = 0 a = 5, b = 7, c = 3 plug these into the formula and solve
*January 19, 2011*

**alg. 2**

2.25 + .15x = 1.50 + .18x solve for x
*January 19, 2011*

**math **

incomplete problem
*January 19, 2011*

**Junior Math**

(t^2-1)/(t^2+7t+10) - (2t+4)/(t^2-t-6) factor the denominators first (t^2-1)/(t+2)t+5) - (2t+4)/(t-3)(t+2) LCD is (t+2)(t+5)(t-3) Now, can you combine over the LCD?
*January 19, 2011*

**ALGEBRA**

x + 2y - 4 = 0 x + 2y = 4 to find x-intercept, y = 0 x + 2(0) = 4 x = 4 x-intercept = (4.0) slope intercept form (which you got right above) y = mx + b for Line A is y = -1/2 x + 2 slope m = -1/2 If two lines are perpendicular, the slope of one is the negative reciprocal of ...
*January 19, 2011*

**math **

5/4 x + 1/8 = 5/8 + x 5/4 x = 4/8 + x multiply both sides by 8, (LCD of 4 and 8) then solve for x
*January 19, 2011*

**math **

you need to use parentheses 5/4x1 divided by 8x = 5/8 + x ? what is "4x1"?
*January 19, 2011*

**Algebra 1**

you're very welcome :)
*January 19, 2011*

**Algebra 1**

f(x) = x^2 + 3 f(5) = 5^2 + 3 f(5) = ? 5^2 = 5 x 5 = 25 f(5) = 5^2 + 3 f(5) = 25 + 3 f(5) = 28
*January 19, 2011*

**Algebra 1**

No, f(5) does not = 10 f(x) = x^2 + 3 f(5) = 5^2 + 3 f(5) = ? 5^2 = ? 5^2 + 3 = ? show me how you got 10
*January 19, 2011*

**Algebra 1**

f(x) = x^2 + 3 f(2) = 2^2 + 3 f(2) = ? 2^2 = 4 f(2) = 2^2 + 3 f(2) = 4 + 3 f(2) = 7 the answer for f(2) is 7 when x = 2, the function value = 7 now, you do the other one
*January 19, 2011*

**Algebra 1**

just plug in the values for x and solve f(x) = x^2 + 3 f(2) = 2^2 + 3 f(2) = ? f(x) = x^2 + 3 f(5) = 5^2 + 3 f(5) = ?
*January 19, 2011*

**math 5th grade**

5/5 * 2/3 = ? 100, 10, 1, 1/10
*January 19, 2011*

**Math**

I = PRT 525 = 3500 *.06 * T Solve for T (time) this is simple interest
*January 19, 2011*

**rectangular area **

you're welcome
*January 19, 2011*

**rectangular area **

A = Lw 240 = 24w w = ?
*January 19, 2011*

**To helper**

I do when they post more than one of the same type problem, but I understand and will only set up for now on
*January 19, 2011*

**Cylinder Height**

r = radius 3r = height (h) SA = 2(pi)(r)(h) + 2(pi)(r^2) SA = 2(pi)(r)(3r) + 2(pi)(r^2) SA = 6(pi)(r^2) + 2(pi)(r^2) SA = 8(pi)r^2 402 = 8(pi)r^2 402 = 8(3.14)r^2 402 = 25.12r^2 16.00318 = r^2 rounding 16.00318 = 16 16 = r^2 r = 4 height = 3r = 3(4) = 12
*January 19, 2011*

**algebra**

|x + 5| > 9 remove absolute value sign x + 5 > 9 x > 4 -x - 5 > 9 -x > 14 when you divide by a negative, reverse the sense of the inequality x < -14 x > 4 x < -14
*January 19, 2011*

**math **

(3xy - 10) - (xy + 5) 3xy - 10 - xy - 5 2xy - 15
*January 19, 2011*

**TO Helper**

I figured those were yours. I'm going through those now (and others). I try to help everyone I can. I don't exactly understand the donut problem either because the equation doesn't seem to work if all donuts were sold. I'm hoping REINY or another math tutor ...
*January 19, 2011*

**TO Helper**

the screen name "Anonymous" belongs to anyone who left the name field blank, so I have no idea which posts are yours. I scan all problems and help where I can. I am not a tutor so I only help on problems I am sure of.
*January 19, 2011*

**MATH ALGEBRA**

2x- 3y + 30 = 0 ? to find x and y intercepts, put equation in the form ax + by = c x intercept = (c/a, 0) y intercept = (0, c/b) to find x-intercept, y = 0 2x - 3(0) = -30 2x = -30 x = -15 x intercept = (c/a, 0) x intercept = (-15, 0) to find y-intercept, x = 0 2(0) - 3y = -30...
*January 19, 2011*

**4th Grade**

2/9 * 36 = 72/9 = 8
*January 19, 2011*

**Algebra 1**

-0.2x < -10 divide both sides by -0.2 when dividing by a neg, reverse the sense of the inequality x > 50
*January 19, 2011*

**algerbra**

yes, you are correct
*January 19, 2011*

**alg. 2**

x^2 - 16x + 4 = 0 x^2 - 16x = -4 (-16^2)/4 = 256/4 = 64 x^2 - 16x + 64 = -4 + 64 x^2 - 16x + 64 = 60 (x - 8)^2 = 60
*January 19, 2011*

**MATH_ALGEBRA**

you're welcome :)
*January 19, 2011*

**MATH_ALGEBRA**

(1) F + 1000P = 32 (2) F + 3500P = 44.50 you have to solve these 2 equations simultaneously By elimination method, to eliminate F, multiply one of the equations by -1 -1(F + 1000P = 32)= -F - 1000P = -32 Now ADD these two equations together -F - 1000P = -32 F + 3500P = 44.50 0...
*January 19, 2011*

**MATH-ALGEBRA**

y = 2/5x - 3 P (0,5) parallel lines have equal slopes slope intercept form of an equation y = mx + b m = slope, b = y-intercept parallel to y = 2/5x - 3, and through P(0, 5) m = 2/5 y intercept = 5 so, the equation would be D. y = 2/5x + 5
*January 19, 2011*

**math**

A = B + 5 B = C + 7 C = THE DIFFERENCE BETWEEN 13 AND 9 D = A - B C = 13 - 9 = 4 B = C + 7 = 4 + 7 = 11 A = B + 5 = 11 + 5 = 16 D = A - B = 16 - 11 = 5 A = 16 B = 11 C = 4 D = 5
*January 19, 2011*

**algebra**

quadratic formula x = (-b +- (sqrt(b^2 - 4ac)))/2a 1) x^2 = 9 x^2 - 9 a = 1, b = 0, c = -9 x = 0 +- (sqrt(0^2 - 4(1)(-9)))/2(1) x = + - (sqrt(0 + 36))/2 x = +- (sqrt36)/2 x = +- 6/2 x = +- 3 x = 3, -3 you try number 2
*January 19, 2011*

**math**

Supplementary angles: two angles whose measures total 180 deg x = 1st angle 3x - 4 = 2nd angle x + 3x - 4 = 180 solve for x to find the angles
*January 19, 2011*

**algebra**

for michele (3x-2)/(5)-2<(2x-3)/(2) multiple both sides by LCD of 10 2(3x - 2) - 20 < 5(2x - 3) 6x - 4 - 20 < 10x - 15 6x - 24 < 10x - 15 -9 < 4x -9/4 < x x > -9/4
*January 19, 2011*

**Math**

if this your problem | 1 - 4x/3 | < 3 remove absolute value sign 1 - 4x/3 < 3 and -1 + 4x/3 < 3 1 - 4x/3 < 3 -4x/3 < 2 -4x < 6 x > -6/4 x > -3/2 -1 + 4x/3 < 3 4x/3 < 4 4x < 12 x < 3 so solution is -3/2 < x < 3
*January 19, 2011*

**math**

from above .03 = value of mixture this should be x + 100 = mixture total number of oz
*January 19, 2011*

**math**

x = number oz of cheaper tea .02x = value of cheaper tea 100 = number oz of better tea .05 * 100 = 5.00 = value of better tea .03 = value of mixture .03(x + 100) = value of mixture .02x + 5.00 = .03(x + 100) .02x + 5.00 = 3.00 + .03x 2.00 = .01x x = 200 200 oz of .02 tea
*January 19, 2011*

**Intermediate Algebra**

V = 2(pi * L^2) or V = 2pi * 2L^2
*January 19, 2011*

**algebra**

(4/(x + 2) - 4/x)/2 = 1 rewrite as (4/(x + 2) - 4/x) * 1/2 = 1 4/(2(x + 2)) - 4/2x = 1 4/(2x + 4) - 4/2x = 1 LCD 2x(2x + 4) (8x - 4(2x + 4))/(2x(2x + 4)) = 1 multiply both sides by 2x(2x + 4) 8x - 4(2x + 4) = 2x(2x + 4) 8x - 8x -16 = 4x^2 + 8x -16 = 4x^2 + 8x 4x^2 + 8x + 16 = ...
*January 19, 2011*

**math!!!**

g(x) = 2x^2 + x - 1 The degree of an equation that has not more than one variable in each term is the exponent of the highest power to which that variable is raised in the equation. b. 2 is the answer
*January 19, 2011*

**math!!!**

h = -6(t - 2.5)^2 + 38.5 plug in a few values for t when t = 1, h = 25 when t = 2, h = 37 when t = 2.5, h = 38 when t = 3, h = 37 you try, what do you think the answer is
*January 19, 2011*

**Algebra**

x = each base vertex angle = 1/2x + 10 solve x + x + 1/2x + 10 = 180
*January 19, 2011*

**Algebra II**

3/x^3 divided by 1/x^2 is this your problem? instead of dividing, invert and multiply 3/x^3 * x^2/1 = 3x^2/x^3 = 3x^-1 = 3/x
*January 19, 2011*

**algebra**

8/26 = m/13 cross multiply means denominator of 1st * numerator of 2nd numerator of 1st * denominator of 2nd 26M = 104 M = 4
*January 19, 2011*

**algebra**

11T/28-T / 4 = 1 rewrite problem 11T/28 - T * 1/4 = 1 11T/(4(28 - t)) = 1 11T/(112 - 4T) = 1 11T = 112 - 4T 15T = 112 T = 7.47
*January 19, 2011*

**advance algebra**

X=smaller number y= bigger number y - x = 192 9x = bigger number 9x - x = 192 8x = 192 x = 24 smaller number bigger number = 9x = 9(24) = 216
*January 19, 2011*

**calculus**

A = Lw 25 = Lw w = 25/L P = 2w + 2L P = 2(25/L) + 2L P = 50/L + 2L
*January 19, 2011*

**Algebra 1**

5x + y = 5x + (3x âˆ’ 11) = 13. where are you getting these problems? there is no way to solve the way you have written this. copy the problem exactly from your book
*January 19, 2011*

**Algebra 1**

no way to solve this
*January 19, 2011*

**math**

cannot solve either as separate equations
*January 18, 2011*

**math**

see your other post
*January 18, 2011*

**ALGEBRA**

4 + x(10 - 8) 4 + 10x - 8x 4 + 2x 2x = -4 x = -2
*January 18, 2011*

**Algebra 1**

5y + 4x = -41 what are you solving for? copy the complete problem from your book there isn't anything to solve for above, unless 5y + 4x = -41 5y = -41 - 4x y = (-41 - 4x)/5 4x = -41 - 5y x = (-41 - 5y)/4 is this what you want there is nothing else to solve for
*January 18, 2011*

**algebra**

don't understand what you need done repost
*January 18, 2011*

**MATH**

881/5 = 176 1/5
*January 18, 2011*

**Algebra 1**

5y + 4x = -41 what are you solving for?
*January 18, 2011*

**Algebra 1**

incomplete problem try again :)
*January 18, 2011*

**Geometry**

a + b + c = 1400 a + b + 600 = 1400 a + b = 800 using Pythagorean theorem a^2 + b^2 = 600^2 a + b = 800 a^2 + b^2 = 600^2 solve simultaneously I got a = 258.579 b = 541.421 or a = 541.421 b = 258.579
*January 18, 2011*

**math(algebra 1)**

2) you are correct x - y = 4 add + y to both sides x - y + y = 4 + y x = 4 + y
*January 18, 2011*

**math(algebra 1)**

1)3x + y = 0, x - y = 4 Pick and equation, and solve for either x or y x - y = 4 x = 4 + y Substitute, x = 4 + y, for x, in the other equation and solve for y x = 4 + y 3x + y = 0 3(4 + y) + y = 0 12 + 3y + y = 0 12 + 4y = 0 4y = -12 y = -3 substitute y = -3 in either equation...
*January 18, 2011*

**Math**

43 heads = 43 animals ( I hope!!) C = number of chickens G = number of goats C + G = 43 2C + 4G = 108 solve simultaneously to find C and G if you are right, C = 32 G = 11
*January 18, 2011*

**algebra**

set up ratio 3/2.5 = x/8 2.5x = 24 x = 9.6 $9.60 for 8
*January 18, 2011*

**5th math**

4 feet = 4 x 12 in/ft = 48 inches from above, this should say write down the most times a factor was used within one factorization
*January 18, 2011*

**5th math**

mean is the average add all the numbers to get the sum divide the sum by the number of numbers you are adding. 33 + 41 + 37 + 27 + 32 = 170 since there are 5 numbers, you divide the sum by 5, to get the mean 170/5 = 34 the mean is 34 LCM 6, 4, 8 LCM is 24 how I was taught ...
*January 18, 2011*

**math**

b. 8th:7th = 2:3
*January 18, 2011*

**Math**

A = 4pir^2 TA = 4(3.14)(5^2) TA = 4(3.14)(25) TA = ?
*January 18, 2011*

**math**

y^2-3y=9 y^2 - 3y = 9 complete the square y^2 - 3y + 9/4 = 9 + 9/4 (y - 3/2)^2 = 45/4 +- (y - 3/2) = 45/4 solve for y y = 3/2 (sqrt5 + 1) y = -3/2 (-1 + sqrt5)
*January 18, 2011*

**algebra 2 and trigonometry**

y^2-3y=9 y^2 - 3y = 9 complete the square y^2 - 3y + 9/4 = 9 + 9/4 (y - 3/2)^2 = 45/4 +- (y - 3/2) = 45/4 solve for y y = 3/2 (sqrt5 + 1) y = -3/2 (-1 + sqrt5)
*January 18, 2011*

**math**

6th - 125 7th - 150 8th- 100 a) ratio 6th:7th 125:150 = 5:6 b) 8th:7th 100:150 = 4:6 = 2:3 c) 6th: total = 125:375 = 5:15 = 1:3 d) 8th:total = 100:375 = 4:15 check your b. ratio 2:3, not 3:2
*January 18, 2011*

**math**

ratio to what? 125 : 150 : 100 5 : 6 : 4
*January 18, 2011*

**Pre-Algebra**

vinegar - pH 3 baking soda - pH 9 9 - 3 = 6 units diff. 10 x 6 = 60x as acidic 10^6 = 1 million, so I can't see how that is right 10^1.77815 = 60 or, 6 x 10^1 = 60 is this a pre-algebra question or science?
*January 18, 2011*

**trig**

restate your question. I don't understand. is this a right triangle? "i have one degree 25" What does this mean? side b = 2.489 * 10^9 ? or 2,489,000,000? post question exactly as written from your book.
*January 18, 2011*

**Pre-Algebra**

:) that's ok. also, you will have better response if posted separate.
*January 18, 2011*

**Pre-Algebra**

vinegar - pH 3 baking soda - pH 9 9 - 3 = 6 units diff. 10 x 6 = 60x as acidic 10^6 = 1 million, so I can't see how that is right 10^1.77815 = 60 I am not a tutor, so you might want to repost this as a separate question
*January 18, 2011*

**Pre-Algebra**

if this is what you mean 1) a^8/a^8 = 1 2) -x^5/-x^4 = x 3) 15/5 * n^9/n = 3n^8 Also, use separate posts for 2 different questions/subjects
*January 18, 2011*

**math**

not enough info to find d and c post the whole problem
*January 18, 2011*

**math 1 digit divisors word problem**

you're welcome :)
*January 18, 2011*

**math 1 digit divisors word problem**

To have a remainder of 8, the divisor must be greater than 8. The only single digit greater than 8 is 9, so 9 must be the divisor. Since 8 is 1 less than 9, the dividend needs to be one less than a multiple of 9. Candidates are values of 3 digits, as the problem states. The ...
*January 18, 2011*

**Emploument English**

true, you are correct
*January 18, 2011*

**Math**

7 1/4 = 29/4 = 58/8 3 3/8 = 27/8 58/8 - 27/8 = 31/8 = 3 7/8
*January 18, 2011*

**math**

see your 1st post
*January 18, 2011*

**math**

7560 = 756. x 10^1 but like Esskay said above, scientific notation would not be used for 7560
*January 18, 2011*

**Sex Ed(English)**

A. 12 weeks
*January 18, 2011*

**math**

x/y - z = rw solve for x x - yz = yrw x = yrw + yz x = y(rw + z) A
*January 18, 2011*

**algebra**

-7 = -1/6x -42 = -x x = 42
*January 18, 2011*

**Sex Ed(English)**

thanks Writeacher, I was not clear :)
*January 18, 2011*

**Sex Ed(English)**

true
*January 18, 2011*

**math**

1/20 of 1.00 = .05 so, 1 nickle or 5 pennies
*January 18, 2011*

**Sex Ed(English)**

VERY correct
*January 18, 2011*

**geometry**

14q + 8 + 8q - 6 = 90 22q + 2 = 90 22q = 88 q = 4 14q + 8 = 64 8q - 6 = 26 64 + 26 = 90
*January 18, 2011*

**Sex Ed(English)**

true
*January 18, 2011*

**math**

5 : x = 25 : 30 5/x = 25/30 cross multiply 150 = 25x x = 6
*January 18, 2011*

**math**

60 x .80 = 48 spent 60 - 48 = 12 left to invest 12 x .80 = 9.60 $9.60
*January 18, 2011*

**Math**

see your 1st post
*January 18, 2011*

**math**

6x - 4y - 3x + y 3x - 3y
*January 18, 2011*

**math**

600 x .06 x 5 = ?
*January 18, 2011*

**MATHAROOO**

think of the midpoint of a line as being the half-point of a line there can only be one point that divides this line into 2 equal parts A--------MP--------B A--------C--------B C is the midpoint of line AB now, you could find the midpoint of segment AC A----MP----C--------B A...
*January 18, 2011*

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