Tuesday

July 7, 2015

July 7, 2015

Total # Posts: 1,869

**math**

Let x = length of side of square thus area of square is A,square = x^2 recall that area of triangle is given by A,triangle = (1/2)b*h where b is the base and h is the height. The base and height of the triangle AMN are equal, which is (1/2)x. Thus, A,triangle = (1/2)*[(1/2)x...
*May 19, 2012*

**algebra.**

let x = width let 6x = length (according to the first statement) recall that area of a rectangle is given by, A = length x width thus, 54 = x * 6x 54 = 6x^2 9 = x^2 x = 3 ft (width) 6x = 18 ft (length) hope this helps~ :)
*May 18, 2012*

**math**

I think you mean, the ratio of the lengths of the smaller and larger cube. Recall that the surface area of a cube is given by SA = 6s^2 where s = length of one side then we get the length of one side of each cube. Larger: 32 = 6s^2 16/3 = s^2 s = 4/sqrt(3) Smaller: 18 = 6s^2 3...
*May 17, 2012*

**math**

I think you mean, the ratio of the lengths of the smaller and larger cube. Recall that the surface area of a cube is given by SA = 6s^2 where s = length of one side then we get the length of one side of each cube. Larger: 32 = 6s^2 16/3 = s^2 s = 4/sqrt(3) Smaller: 18 = 6s^2 3...
*May 17, 2012*

**chemistry**

first we write the chemical equation: C6H6 + O2 -> CO2 + H2O then we balance this: 2C6H6 + 15O2 -> 12CO2 + 6H2O then we solve for the molecular weight of C6H6. Note that mass of C = 12 and H = 1. thus, 6*12 + 6*1 = 78 g/mol then we get the number of moles of C6H6 used in...
*May 16, 2012*

**chemistry**

first we write the chemical equation: C6H6 + O2 -> CO2 + H2O then we balance this: 2C6H6 + 15O2 -> 12CO2 + 6H2O then we solve for the molecular weight of C6H6. Note that mass of C = 12 and H = 1. thus, 6*12 + 6*1 = 78 g/mol then we get the number of moles of C6H6 used in...
*May 16, 2012*

**Geometry**

*i mean 56 degrees.
*May 16, 2012*

**Geometry**

it's difficult to show solutions (because diagrams are required), but my answer is 58 degrees.
*May 16, 2012*

**Chemistry**

H2 + O2 -> H2O balancing this, 2H2 + O2 -> 2H2O then we find the limiting reactant. To determine the limiting reactant, we get the moles of product from the given moles of reactant. Whichever yield less number of moles of product is the limiting reactant. thus, 4 mol H2...
*May 16, 2012*

**Math**

270 degrees
*May 15, 2012*

**math**

(2x-5)/6 = (4x+1)/2 we must make the denominators same. to do this, we multiply the numerator & denominator of the term on the right by 3: (2x-5)/6 = 3(4x+1)/6 then we equate the numerators: 2x - 5 = 12x + 3 2x - 12x = 5 + 3 -10x = 8 x = -4/5
*May 15, 2012*

**math 8th grade**

multiply the given dimensions: 4 * 12 = 36 5 * 11 = 55 7 * 9 = 63 8 * 8 = 64 therefore, it's 8 in x 8 in.
*May 14, 2012*

**math 8th grade**

recall that area of rectangle (shape of the photo) is given by A = length * width substituting, A (original) = 3 * 5 = 15 if we triple both length and width, new length = 3 * 5 = 15 new width = 3 * 3 = 9 thus, A (new) = 15 * 9 = 105 to get the ratio, we divide the new area by ...
*May 14, 2012*

**algebra**

to get the inverse, we just interchange the variables x and y, and solve for y> Recall that we can also substitute f(x) as y: y = 5x^3 - 9 then we interchange x and y: x = 5y^3 - 9 and we solve for y: x + 9 = 5y^3 (x + 9)/5 = y^3 y = [(x + 9)/5]^(1/3) then finally we ...
*May 14, 2012*

**algebra**

we just substitute the value of x for x + h : f(x) = x^2 - 2 f(x+h) = (x+h)^2 - 2 if we want to expand it, f(x+h) = x^2 + 2xh + h^2 - 2 hope this helps~ :)
*May 14, 2012*

**math**

3/4 or 75% of the people were students. thus, 100 - 75 = 25% of the people were parents (moms & dads) of these, there are 3 dads to 4 moms. Note that the total is 3 + 4 = 7 parents. We can say that 4/7 or 57.14% of the parents are moms. 25% of this is the percent mom: (25% * ...
*May 14, 2012*

**Algebra 2**

(a) log(base10) of 8 log(base10) of 8 can be rewritten as = log(base10) of (2^3) = 3*[ log(base10) of 2 ] = 3*0.301 = 0.903 (b) log(base10) of 12 log(base10) of 12 can be rewritten as = log(base10) of (2*2*3) = log(base10) of 2 + log(base10) of 2 + log(base10) of 3 = 0.301 + 0...
*May 9, 2012*

**math**

Two ways: (1) Plot y = 3x - 5 and check if it passes through the point (4,7). (2) Substitute the x-coordinate of the point given (which is 4) to the equation, and solve for y. If the resulting y is equal to 7, it is a solution to the equation. Let's do #2 to check: y = 3x...
*May 8, 2012*

**MATH**

6 1/2 - 4 7/10 first, we convert this to improper fraction (numerator is greater than denominator). We do this be multiplying the whole number to the denominator then add to the numerator: (6*2 + 1)/2 - (4*10 + 7)/10 (12 + 1)/2 - (40 + 7)/10 13/2 - 47/10 Then, we find the ...
*May 8, 2012*

**chemistry**

For an ideal gas, (P1*V1)/(T1) = (P2*V2)/(T2) where P1, V1 and T1 are the pressure, volume and temp at initial conditions, and P2, V2 and T2 are the final conditions Substituting, (104*2.5)/270 = (95*5.3)/(T2) now solve for T2.
*May 8, 2012*

**math**

(4 + 1 - 3)/ 2 = 1
*May 8, 2012*

**ALGEBRA1**

Let D = amount of money needed to reach $635 therefore, 365 + D = 635 D = 635 - 365 D = $270
*May 8, 2012*

**math**

3/8 and 18/48
*May 8, 2012*

**maths**

f(x) = (x-1)/(2x^2) - (x-3) #1. f(2) = (2-1)/(2*2^2) - (2-3) f(2) = 1/(8) - (-1) f(2) = 1/8 + 1 f(2) = 9/8 #2. f(x) = (x-1)/(2x^2) - (x-3) 0 = (x-1)/(2x^2) - (x-3) x - 3 = (x-1)/(2x^2) (2x^2)(x-3) = x - 1 2x^3 - 6x^2 - x + 1 = 0 x = -0.459 x = 0.350 x = 3.11 #3. domain is all ...
*February 18, 2012*

**maths**

f(x) = (x-1)/(2x^2) - (x-3) #1. f(2) = (2-1)/(2*2^2) - (2-3) f(2) = 1/(8) - (-1) f(2) = 1/8 + 1 f(2) = 9/8 #2. f(x) = (x-1)/(2x^2) - (x-3) 0 = (x-1)/(2x^2) - (x-3) x - 3 = (x-1)/(2x^2) (2x^2)(x-3) = x - 1 2x^3 - 6x^2 - x + 1 = 0 x = -0.459 x = 0.350 x = 3.11 #3. domain is all ...
*February 18, 2012*

**math**

1/2*9.8*4^2 note that four exponent 2 means 4*4, which is equal to 16. thus, 1/2*9.8*16 you just type this in a calculator. the answer you'll get is 78.4
*February 10, 2012*

**Algebra**

(a) slope = m = 1 (b) m = 0 (c) m = undefined
*February 10, 2012*

**math**

1 - 7/10 = 3/10
*February 9, 2012*

**algebra**

let x = gallons needed 120 : 7 = 180 : x , or 120/7 = 180/x x/7 = 180/120 x = 7*3/2 x = 10.5 gallons
*February 7, 2012*

**intermediate algebra**

let x = first angle let 90-x = second angle then we set up the equation: 8x = (90-x) 8x + x = 90 9x = 90 x = 10 degrees 90-x = 80 degrees hope this helps~ :)
*February 4, 2012*

**intermediate algebra**

let x = first angle let 180-x = second angle then we set up the equation, 14x = 180-x 14x + x = 180 15x = 180 x = 12 degrees 180-x = 168 degrees hope this helps~ :)
*February 4, 2012*

**intermediate algebra**

let x = first angle let 180-x = second angle then we set up the equation, 14x = 180-x 14x + x = 180 15x = 180 x = 12 degrees 180-x = 168 degrees hope this helps~ :)
*February 4, 2012*

**Math**

(-5^2+(x+2))/-4-(-6) = -10 (-25 + (x+2))/(-4+6) = -10 (-25 + (x+2))/2 = -10 -25 + x + 2 = -20 -23 + x = -20 x = -20 + 23 x = 3 so, it's letter (E). hope this helps~ :)
*February 4, 2012*

**algebra**

to do this, you just divide 1350 by 45. 1350/45 = ?
*February 1, 2012*

**Math**

3sqrt(2) + 5sqrt(4) + 8sqrt(2) - 2sqrt(6) 11sqrt(2) + 5*2 - 2sqrt(6) 10 + 11sqrt(2) - 2sqrt(6) hope this helps~ :)
*February 1, 2012*

**Algebra**

if we're solving for x, we will separate each parenthesis term and equate them to zero. there are two parenthesis terms in the problem: the (2x+1) and the (x-4). now we equate each to zero, and solve for x. 2x + 1 = 0 2x = -1 x = -1/2 (the first root) x - 4 = 0 x = 4 (the ...
*February 1, 2012*

**Additional maths**

let a = one root let (1/3)a = the other root recall that quadratic equation can also be written as, x^2 - (sum of roots)x + (product of products) = 0 thus, product of roots = -12 -12 = (1/3)a^2 -36 = a^2 a = 6i (1/3)a = 2i *note that i = sqrt(-1) , which is an imaginary number...
*February 1, 2012*

**Additional maths**

another solution: recall that a quadratic equation can also be rewritten as, x^2 - (sum of roots)x + (product of roots) = 0 where the numerical coeff of x = -(sum of roots), and constant = (product of roots) therefore, sum of roots = -2 + (-3) = -5 equating this to the ...
*February 1, 2012*

**Measurement:: Metric Units**

to do this, you get the perimeter of the table. recall that the perimeter of a rectangle is given by, P = 2*L + 2*W where L = length W = width substituting, P = 2(1.75) + 2(1) P = 3.5 + 2 P = 5.5 meters then we convert this to centimeters. note that 1 meter = 100 centimeters ...
*January 31, 2012*

**Trig**

2cos(x-π/6) = 1 let y = (x-π/6) cos(y) = 1/2 y = π/3, -π/3 substituting back the value of y, (x - π/6) = π/3 x = π/2 (x - π/6) = -π/3 x = -π/6 or 11π/12 hope this helps~ :)
*January 31, 2012*

**Precalculus**

note: to subtract log with equal base (in this case, the base is 10) we just divide the terms inside the log. thus, 4log(2) + log(6) - log(3) 4log(2) + log(6/3) 4log(2) + log(2) finally we add: 5log(2) or log(32) hope this helps~ :)
*January 30, 2012*

**Chemistry**

sodium propionate, since ion-dipole interaction between water & sodium propionate is stronger than hydrogen bonding of water & propionic acid
*January 30, 2012*

**Chemistry**

Glucose, since it has -OH groups and thus can do hydrogen bonding with water. Unlike cyclohexane, which does not have -OH groups.
*January 30, 2012*

**Math**

...limit as x approaches what value?
*January 30, 2012*

**MATH**

note that when we substitute 59 to the expression in numerator and denominator, (sqrt(x+5)-8)/(x-59) (sqrt(59+5)-8)/(59-59) (sqrt(64)-8)/0 8-8/0 0/0 thus we use L'hopital's Rule: we separately get the derivative of numerator and denominator. thus, (num) = (sqrt(x+5)-8...
*January 30, 2012*

**chemistry**

the only chemical change there is the burning of wood.
*January 29, 2012*

**chemistry**

Burning of gasoline (combustion) note that phase changes are physical changes only. hope this helps~ :)
*January 29, 2012*

**Precalculus**

yes.
*January 29, 2012*

**Precalculus**

ahh i think you mean cube root for that ^3sqrt(64). if so, log(base4) of (cuberoot(64)) note that we can rewrite cuberoot(64) as cuberoot(4^3) which is equal 4. log(base4) of 4 = 1 hope this helps~ :)
*January 29, 2012*

**Algebra II**

#1 (2x^(-3)y^(2))^(-2) (2)^(-2) * x^(-3*(-2)) * y^(2*(-2)) (1/4) * x^(6) * (1/y^4) (x^6)/(4y^4) #2 (z^4)/(2z^3) z/2 hope this helps~ :)
*January 29, 2012*

**algebra**

recall that the slope, m, is given by the equation, m = (y2 - y1)/(x2 - x1) where (x1, y1) = coordinates of first point (x2, y2) = coordinates of second point substituting, m = (1 - 6)/(-3 - 5) m = (-5)/(-8) m = 5/8 hope this helps~ :)
*January 29, 2012*

**algebra**

4x - 15/2 = 1/2 first we transpose all terms with no variable x to the right side of equation. that is, we transpose - 15/2 to the right. note that when we transpose, we change the sign to its opposite. thus is becomes positive: 4x = 1/2 + 15/2 when adding fractions, we first ...
*January 29, 2012*

**Precalculus**

Bosnian's answer is correct. I'll just show how to get 0. 6^(x+3) - 6^x = 215 recall that when multiplying exponents of the same base, we add their exponents. for example, x^2 * x^4 = x^(2+4) = x^6 now we do the opposite to 6^(x+3). we'll separate this into, 6(x+3...
*January 29, 2012*

**math**

SA of sphere = 4*pi*r^2 154 = 4*3.14*r^2 r = 3.502 V of sphere = (4/3)*pi*r^3 V = (4/3)*3.14*(3.502)^3 V = 170.81 cm^3 hope this helps~ :)
*January 29, 2012*

**Precalculus**

5^(2-x) = 1/125 note that we can rewrite 125 as 5^3 thus, we replace the term on the right side of equation: 5^(2-x) = 1/(5^3) note also that we can still rewrite 1/(5^3) as 5^(-3). thus, 5^(2-x) = 5^(-3) now that the bases (the 5) are equal, we equate their exponents: 2 - x...
*January 28, 2012*

**math**

..i think you forgot to type the expressions..
*January 28, 2012*

**algebra**

-3|6n-2| +5 = 8 case I: absolute value is positive. thus, -3(6n-2) + 5 = 8 -3(6n-2) = 8 - 5 -3(6n-2) = 3 6n - 2 = -1 6n = 1 n = 1/6 case II: absolute value is negative. thus, -3[-(6n-2)] + 5 = 8 3(6n-2) + 5 = 8 3(6n-2) = 3 6n - 2 = 1 6n = 3 n = 1/2 NOTE: these are not yet the ...
*January 28, 2012*

**trig**

oops i forgot the +/- sign: cos (theta) = +/- sqrt(3) / 2 theta = +/- arccos [ sqrt(3) / 2 ] theta = π/6 , 11π/6 , 5π/6 , 7π/6
*January 22, 2012*

**trig**

4cos^2 theta = 3 cos^2 theta = 3/4 cos (theta) = sqrt(3) / 2 theta = arccos [ sqrt(3) / 2 ] theta = π/6 , 11π/6 hope this helps~ XD
*January 22, 2012*

**math problem**

we start performing the operation inside the parenthesis, then going outside of the parenthesis: 8[-73-(-61-28)] 8[-73-(-89)] 8[-73+89] 8[16] 128 hope this helps~ XD
*January 22, 2012*

**math**

5x^2 = 9x + 2 first we transpose all terms to the left side of the equation: 5x^2 - 9x - 2 = 0 then we factor (since it's factorable): (5x + 1)(x - 2) = 0 5x + 1 = 0 5x = -1 x = -1/5 x - 2 = 0 x = 2 therefore, x = -1/5 and x = 2 hope this helps~ XD
*January 22, 2012*

**physics**

what is the distance, given the velocity that has magnitude 1.2 m/s within 40 min.?
*December 29, 2011*

**math **

(a) miles per hour 26.2 / 2.5 = ? (b) miles per minute: first convert 2.5 hours to minutes. 2.5 * 60 = 150 min 26.2 / 150 = ? (c) minutes per mile: just get the reciprocal of your answer in (b)
*November 10, 2011*

**Math**

when you want a variable to be squared you use this symbol, ^ , which means "raised to". in the problem, s = ut + (1/2)*(g*t^2) s - (1/2)*(g*t^2) = ut [ s - (1/2)*(g*t^2) ]/t = u, or u = [ s - (1/2)*(g*t^2) ]/t
*November 10, 2011*

**math**

(a) to write this in standard form, you use completing the square for both variables x and y. x^2 + y^2 + 8y - 65 = 0 x^2 + (y^2 + 8y + 16) - 65 - 16 = 0 x^2 + (y + 4)^2 - 81 = 0 x^2 + (y + 4)^2 = 81 x^2 + (y + 4)^2 = 9^2 (b) Center: C(0, -4) Radius = 9 to get intercepts, x-...
*November 10, 2011*

**Algebra**

#1. (5y-3)/3 - (2y-1)/2=1 [(5y-3)/3 - (2y-1)/2 = 1]*6 2(5y-3) - 3(2y-1) = 6 10y - 6 - 6y + 3 = 6 4y - 3 = 6 4y = 9 y = 9/4 #2. -5(-2(x-1) - (x+1)) = 9 -5(-2x + 2 - x - 1) = 9 -5(-3x + 1) = 9 15x - 5 = 9 15x = 14 x = 14/15 #3. .08n-.09(n-2)=5-.08 0.08n - 0.09n + 0.18 = 4.92 -0....
*November 1, 2011*

**math**

let x = larger number let y = smaller number then we set up the equations. since their difference is 25, x - y = 25 : equation (1) since the smaller is 1/6 of the larger, y = (1/6)x : equation (2) we substitute equation (2) to equation (1): x - (1/6)x = 25 (5/6)x = 25 x = 25*6...
*November 1, 2011*

**algebra**

to get the x-intercept, we set y = 0 and solve for x: 2x + 3y = 15 2x + 3(0) = 15 2x = 15 x = 7.5 thus, x-intercept is at (7.5, 0) to get y-intercept, we set x = 0 and solve for y: 2x + 3y = 15 2(0) + 3y = 15 3y = 15 y = 5 thus, y-intercept is at (0, 5) to plot this, locate ...
*November 1, 2011*

**algebra**

to get the x-intercept, we set y = 0 and solve for x: 2x + 3y = 15 2x + 3(0) = 15 2x = 15 x = 7.5 thus, x-intercept is at (7.5, 0) to get y-intercept, we set x = 0 and solve for y: 2x + 3y = 15 2(0) + 3y = 15 3y = 15 y = 5 thus, y-intercept is at (0, 5) to plot this, locate ...
*November 1, 2011*

**algebra**

to get the y-coordinates, we just substitute the value of the x-coordinates given. at x = -2: y = 4x + 2 y = 4(-2) + 2 y = -8 + 2 y = -6 thus (-2, -6) at x = 1: y = 4x + 2 y = 4(1) + 2 y = 4 + 2 y = 6 thus (1, 6) at x = -3: y = 4x + 2 y = 4(-3) + 2 y = -12 + 2 y = -10 thus (-3...
*November 1, 2011*

**algebra**

v = (1/3)t + 12 to plot this, the easiest way (for me) is to get the x- and y-intercept (in this case, the x-axis = t-axis, and y-axis = v-axis). to get t-intercept, we let v = 0 and solve for t: v = (1/3)t + 12 0 = (1/3)t + 12 -(1/3)t = 12 t = -36 thus t-intercept is at (-36...
*November 1, 2011*

**chemistry**

cations - ions with positive charges (ex: Na+ , Ca2+) anions - ions with negative charges (ex: Cl- , OH-)
*October 29, 2011*

**chemistry**

Molarity = moles of solute / Liters of solution moles = mass given / molar mass substituting, Molarity = (5.20/36.46)/3 Molarity = 0.0475 M hope this helps~ :)
*October 29, 2011*

**math**

let x = hypotenuse let y = the length of the other leg we have two equations, two unknowns here. recall that perimeter of a triangle is just, P = a + b + c substituting, 38.7 = x + y + 12.5 x + y = 38.7 - 12.5 x + y = 26.2 x = 26.2 - y : : equation (1) recall that for any ...
*October 29, 2011*

**Algebra 1**

we are solving for x. To get x alone, we need to isolate the term with x only to one side of the equation. In this case, let's isolate the term with x (which is 0.3x) on the right side, thus we will transpose the -9.1 on the left side. To do this, the its sign must be ...
*October 29, 2011*

**math**

the surface area of a rectangular solid is given by SA = 2(LW + LH + WH) where W = width, L = length, H = height if we double all the dimensions, SA' = 2[(2L)(2W) + (2L)(2H) + (2W)(2H)] SA' = 2[4LW + 4LH + 4WH] SA' = 2*4[LW + LH + WH] SA' = 4*SA therefore, what...
*October 28, 2011*

**algebra**

15/30 * 100 = 50%
*October 28, 2011*

**algebra**

let x = number note that 80% is also equal to 0.8 88 = (0.8)x x = 88/0.8 x = 110
*October 28, 2011*

**Algebra**

we just substitute the value of x and y to the equation: 3x + c = 5y 3(-4) + c = 5(0) -12 + c = 0 c = 12
*October 28, 2011*

**algebra**

-9 - (-3) -9 + 3 -6
*October 28, 2011*

**Algebra**

let x = number of multiple questions she got right. then we set up the equation and solve for x: (4-1)*7 + 3x = 78 21 + 3x = 78 3x = 57 x = 19 hope this helps~ :)
*October 28, 2011*

**Geometry**

it's difficult to explain it here because you need to draw to see it clearly. i'll provide a link to for you to see the drawing (though it's little unclear :P) first you draw the circle and the parallel chords. connect the center of circle to one of the endpoints ...
*October 28, 2011*

**science**

recall that heat absorbed released is given by Q = mc*(T2 - T1) where m = mass (in g) c = specific heat capacity (in J/g-k) T = temperature (in C or K) *note: Q is (+) when heat is absorbed and (-) when heat is released. we're looking for (T2 - T1) here. substituting, ...
*October 28, 2011*

**science**

recall that heat absorbed released is given by Q = mc*(T2 - T1) where m = mass (in g) c = specific heat capacity (in J/g-k) T = temperature (in C or K) *note: Q is (+) when heat is absorbed and (-) when heat is released. substituting, Q = (480)*(0.97)*(234 - 22) Q = 98707 J = ...
*October 28, 2011*

**trignometry**

the derivative of cot x = -csc^2 x the derivative of tan x = sec^2 x .. and you do this: [ low*d(high) - high*d(low) ] / (low)^2 where high = numerator low = denominator d(high) and d(low) = respective derivatives
*October 28, 2011*

**Math**

add the two equations to get y alone: y = -3/4x + 1/4 y = 3/4x - 3/4 -------------------------- 2y = -2/4 y = -1/4 then substitute this value of y to either of the equations and solve for x. in this case let's substitute this to the first give equation: y = -3/4x + 1/4 -1/...
*October 28, 2011*

**derivatives**

[ low*d(high) - high*d(low) ] / (low)^2 where high = numerator low = denominator d(high) and d(low) = respective derivatives first recall that the derivative of csc x = -(cot x)(csc x) therefore, [ x*(-6 (cot x)(csc x)) - 6 (csc x) ]/x^2 or -(6 csc x)*(x(cot x) + 1)/x^2 hope ...
*October 28, 2011*

**math **

but we don't se the diagram..
*October 28, 2011*

**HELP**

(3v - 9)/(3 - v) [3(v-3)]/[-1(v-3)] -3
*October 28, 2011*

**HELP**

it must be 4a^7t^2/3 (remove the t^2 from the denominator)
*October 27, 2011*

**chemistry**

to get this, we just divide the final volume by the initial volume: 55.5 / 7.4 = ?
*October 27, 2011*

**Algebra**

at x = 2 8y + 7*2 = -25 8y = -25 - 14 y = -39/8 at x = 4 8y + 7*4 = -25 8y = -25 - 28 y = -53/8 at x = 6 8y + 7*6 = -25 8y = -25 - 42 y = -67/8 thus it's (2, -39/8), (4, -53/8), (6, -67/8) hope this helps~ :)
*October 27, 2011*

**Precalc**

x^2 - 4x + 6 first, yo only focus on the first two terms: x^2 and -4x. let's put a parenthesis to enclose them: (x^2 - 4x) + 6 now, to complete the square, what we do is get the half of b (b is the numerical coefficient of x, and in this case, is -4) and we square it: (-4/...
*October 27, 2011*

**Algebra**

.25(250)+250/250 62.5 + 1 63.5
*October 27, 2011*

**Algebra**

yes.
*October 27, 2011*

**Alg. II**

54 + 16x^3 2(8x^3 + 27) then recall the how we factor sum of two cubes: a^3 + b^3 = (a+b)(a^2 - ab + b^2) therefore, 2(2x+3)(4x^2 - 6x + 9) hope this helps~ :)
*October 27, 2011*

**Trig/Algebra II**

(3+√2)/(3-√2) (3+√2)*(3+√2)/(3-√2)(3+√2) (9 + 6√2 + 2)/(9-2) (11 + 6√2) / 7 it's A.
*October 27, 2011*

**physics**

Q = Pt = mc(T2-T1) where Q = heat/energy (in J) P = Power (in W) t = time (in s) m = mass (in kg) c = specific heat capacity (in J/kg-K) = 4.184 J/kg-K for water T = temperature (in C or K) we're solving for mass. substituting, (7.5*1000)*(3*24*60*60) = m*4.184*(70-18) ...
*October 27, 2011*

**Int. Algebra**

A. substitute r = 1800 to the equation: 1800 = 200*sqrt(t+3) - 600 2400 = 200*sqrt(t+3) 12 = sqrt(t+3) 144 = t+3 t = 141 hours B. recall that area of circle is just A = pi*r^2. thus we square both sides and multiply by pi: r(t) = 200*sqrt(t+3) - 600 r^2 = 40000(t+3) - 240000*...
*October 26, 2011*

**math**

3x + 5y = 55 the condition that must be satisfied is that x, y and z must all be multiples of 5. first, we guess value of x (which must be a multiple of 5). And then we solve for corresponding value of y. Note that the value of y calculated must also be a multiply of 5. If it...
*October 26, 2011*