Monday

July 25, 2016
Total # Posts: 1,995

**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*

**algebra**

let x = cost of a record 20x + 0.89
*October 26, 2011*

**science**

the conversion factor for calories (cal) and joules (J) is: 1 cal = 4.184 J now to convert calories to joules, we multiply 4000 cal by 4.184 J / 1 cal so that the cal unit will be canceled, and the numerator will have the unit J: 4000 cal * 4.184 J / cal = 16736 J hope this ...
*October 26, 2011*

**math**

recall PEMDAS: Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction. this is the order which must be followed when performing multiple operations. therefore, (3 x 6) + 12/2 - 8 18 + 12/2 - 8 18 + 6 - 8 24 - 8 16 hope this helps~ :)
*October 26, 2011*

**math**

that's right. you subtract 13 to both sides of equation to get x alone. x + 13 - 13 = 25 - 13 x = 12
*October 26, 2011*

**Math**

recall that rate is just distance travelled over time, or v = d/t let x = time required for them to meet thus since they are paddling toward each other, we add their rates and multiply it by the time required and equate this distance to 33: (4 + 7)x = 33 11x = 33 x = 3 hours ...
*October 26, 2011*

**MATH 7th GRADE**

everything is correct except #7. it must be 8 1/4 = 33/4 1 3/8 = 11/8 (33/4) / (11/8) 33/4 * 8/11 = 6 hope this helps~ :)
*October 26, 2011*

**Calculus**

recall that to get the derivative of 2 functions of x which are multiplied, we do this: d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x) therefore, d/dx 3x((x+1)^1/2) = 3(x+1)^(1/2) + (3x)*(1/2)*(x+1)^(-1/2) hope this helps~ :)
*October 26, 2011*

**math**

let x = number of chicken let y = number of goat let 100 - (x+y) = number of sheep 0.5x + 3.5y + 5[100-(x+y)] = 100 0.5x + 3.5y + 500 - 5x - 5y = 100 -4.5x - 1.5y = -400 9x + 3y = 800 there are actually many possible answers here since you got only 1 equation and 2 unknowns. ...
*October 26, 2011*

**Math**

a, c, f
*October 25, 2011*

**Chemistry **

assuming ideal gas, we can use the formula for constant pressure: V1/T1 = V2/T2 where V = volume T = temperature (in Kelvin) note that to convert to Kelvin, K = C + 273 substituting, 39.5 / (25 + 273) = 7.7 / T2 now solve for T2. subtract 273 to convert it to deg C units
*October 25, 2011*

**Math**

we subtract the price of magazine from 59.99 to get the total amount of the 3 books: 59.99 - 1.99 = 58.00 (cost of 3 books) thus, we divide this by 3 to get the price of each book: 58.00 / 3 = 19.33 hope this helps~ :)
*October 24, 2011*

**math**

recall that speed is distance travelled over time. let x = speed of the other car. then we set-up the equation. when you draw the directions they're travelling, you will form a right triangle. thus recall that in a right triangle, the hypotenuse (in this the the distance ...
*October 24, 2011*

**math**

recall that speed is distance travelled over time. let x = speed of the other car. then we set-up the equation. when you draw the directions they're travelling, you will form a right triangle. thus recall that in a right triangle, the hypotenuse (in this the the distance ...
*October 24, 2011*

**geometry**

but we need to see the figure..
*October 24, 2011*

**math**

recall that speed is distance travelled over time, or v = d/t substituting, v = 10/30 v = 0.33 km/min hope this helps~ :)
*October 24, 2011*

**Trigonometry**

sec (π - x) = -sec (x) when we prove, we only manipulate the left side. first, recall that sec (x) = 1/cos x. thus we can replace the the term on the left side of equation by: 1 / cos (π - x) then recall the sum/difference identity for cosine: cos(A−B) = cos A ...
*October 23, 2011*

**Math **

to get this, we get the derivative of h(t) with respect to t, and equate it to zero: h(t) = 2 + 28t - 4.9t^2 0 = 28 - 9.8t 9.8t = 28 t = 2.86 s *note that the derivative of a function is the slope of the tangent line at the given point (in this problem, the point referred is ...
*October 23, 2011*

**Math**

first we represent the unknowns using variables: let x = length let y = width recall that perimeter of rectangle is just, P = 2x + 2y since P is given (which is 400), we can get an expression of width in terms of length: 400 = 2x + 2y 400 - 2x = 2y y = 200 - x then to get the ...
*October 23, 2011*

**Math **

first we solve for hypotenuse: to get the hypotenuse we use pythagorean theorem: c^2 = a^2 + b^2 where c = hypotenuse of the right triangle a and b = legs of right triangle substituting, c^2 = 6^2 + 3^2 c^2 = 36 + 9 c^2 = 45 c = 3*sqrt(5) therefore the perimieter is: P = a + b...
*October 23, 2011*

**Math**

C. because 5/25 = 1/5, while 4/24 = 1/6
*October 23, 2011*

**College Algebra**

y = (kx^3)/z substituting the given values to find the constant k, 16 = k(4^3)/8 16 = k(64)/8 16 = 8k k = 2 therefore, y = (2x^3)/z if x = 9 and z = 6, y = 2*(9^3)/6 y = 243 hope this helps~ :)
*October 23, 2011*

**calculus**

lim (5x^3 + 7x^2 - 9x + 14) as x = 7 to do this just substitute x = 7: 5*(7^3) + 7*(7^2) - 9*7 + 14 = 2009 hope this helps~ :)
*October 23, 2011*

**geomatry**

<d is always less than <e.
*October 23, 2011*

**chemistry**

assuming the gas is ideal, we can use the ideal gas law: PV = nRT where P = pressure (in atm) V = volume (in L) n = number of moles R = gas constant = 0.0821 L-atm/mol-K T = temperature (in K) note that n = mass given / molar mass. The molar mass of nitrogen (diatomic gas) is ...
*October 23, 2011*

**Algebra 2**

first represent the unknown using variable: let x = time required so that they will have the same number of rows. recall that rate = quantity/time, or quantity = rate*time in this problem, quantity refers to number of rows knitted. to set-up the equation, what we only know is ...
*October 23, 2011*

**maths**

you mean how did this happen: sqrt(245) = 7*sqrt(5) if so, first you factor out 245: 245 = 5 * 49 = 5 * 7 * 7 = 5 * 7^2 therefore, sqrt(5 * 7^2) we can also rewrite this as sqrt(5) * sqrt(7^2) we can take out the seven since the sqrt(7^2) is equal to 7, and leave the 5 inside...
*October 23, 2011*

**reposting maths qn as mistake in quantity**

i couldn't really think of a solution which is non-algebraic. but i just tried to simplify things in this solution: 16 Forks + 12 Spoons = 106.40 therefore, if we get the half of both the number of forks and spoons, the price will also be half: 8 Forks + 6 Spoons = 53.20 ...
*October 23, 2011*

**Math**

we'll work backwards: x = 3 and x = -2 x - 3 = 0 and x + 2 = 0 then we multiply both these functions of x: (x - 3)(x + 2) = 0 x^2 - 3x + 2x - 6 = 0 x^2 - x - 6 = 0 hope this helps~ :)
*October 23, 2011*

**Algebra**

2e^2 = 3h e^2 = (3/2)*h e = sqrt((3/2)*h)
*October 23, 2011*

**Pre Calculus**

f(x) = 1/sqrt(x+1) we can rewrite this as y = 1/sqrt(x+1) to get the inverse function, we replace x by y, and y by x, and solve for y: x = 1/sqrt(y+1) sqrt(y+1) = 1/x y + 1 = 1 / x^2 y = (1/(x^2)) - 1 f'(x) = (1/(x^2)) - 1 for #2, we do the same: f(x) = 1 - x^2 y = 1 - x^2...
*October 23, 2011*

**Physics**

i suggest you read "Twin Paradox"
*October 23, 2011*

**Science**

recall that heat absorbed or released is given by Q = mc(T2 - T1) where m = mass (in kg) c = specific heat capacity (in J/kg-K) T = temperature (in C or K) *note: Q is (+) when heat is absorbed, and (-) when heat is released. #1. we just substitute. the c for water is equal to...
*October 23, 2011*

**Math**

recall that the formula for discriminant: D = b^2 - 4ac if D < 0 : two imaginary roots D = 0 : one root D > 0 : two real roots thus, given the equation, we can substitute the values of a, b and c in the discriminant, which must be > 0: (2k - 3)^2 - 4(4)(1) > 0 4k^2...
*October 23, 2011*

**Adv algebra**

there is only one number which satisfies "the square of itself is equal to itself": it is 1. (0 is also the other but it is stated that q cannot be equal to 0) therefore, (x-1)(x-1) = x^2 + 2x + 1 p = 2 q = 1 hope this helps~ :)
*October 22, 2011*

**Maths Calculus**

we can rewrite this as sqrt(x^6)*sqrt(3x^5 + 1)*sin(7x) sqrt(3x^11 + x^6)*sin(7x) (3x^11 + x^6)^(1/2) * sin(7x) then recall that derivative of two functions of x that are multiplied is equal to, d/dx ( f(x)*g(x) ) = f'(x)g(x) + f(x)g'(x) therefore, (1/2)*(33x^10 + 6x^5...
*October 22, 2011*

**math**

let x = cost of bag let y = cost of shoes let 160 - (x+y) = cost of watch set up equations: (1) x = y + 15 (2) 160 - (x+y) = x - 20 solve simultaneously. we can either use elimination or substitution, but for this problem let's just use substitution. we substitute equation...
*October 22, 2011*

**science**

(a) recall that potential energy is stored energy, and is given by the formula: PE = mgh (units in Joules) where m = mass (in kg) g = acceleration due to gravity = 9.8 m/s^2 h = height (in m) if the reference position is at the floor (that is, h = 0), the PE is equal to PE = 4...
*October 22, 2011*

**Math **

we substitute the given values and solve for a. at x = 2, f(x) = 3: 3 = a(2^2) - 6(2) - 7 3 = 4a - 12 - 7 3 = 4a - 19 3 + 19 = 4a 22 = 4a a = 5.5 substituting back, f(x) = 5.5x^2 - 6x - 7 hope this helps~ :)
*October 21, 2011*

**Math**

let n = number (n/64) + 48
*October 21, 2011*

**MATH (REPLY FAST)**

4 - 3|m + 2| > -14 there are two cases here. Case I: |m + 2| = m + 2 4 - 3(m+2) > -14 4 - 3m - 6 > -14 -3m - 2 > -14 -3m > -12 m < 4 Case II: |m + 2| = -(m + 2) 4 - (-3(m+2)) > -14 4 + 3m + 6 > -14 3m + 10 > -14 3m > -24 m > -8 therefore, all ...
*October 21, 2011*

**chemistry**

one mole of solid Cobalt (II) Chloride reacts with 2 moles of Hydrogen Fluoride to yield one mole of solid Cobalt (II) Fluoride and two moles Hydrogen Chloride. *for #2 i think that's Pb not P6, and KCl, not KL. one mole of aqueous Lead (II) Nitrate reacts with two moles ...
*October 21, 2011*

**math**

let x = amount of money of Lexi let 250 - x = amount of amount of Marie then set up the equation: x - (2/5)x = 250 - x - 40 (3/5)x = 210 - x (3/5)x + x = 210 (8/5)x = 210 x = 210*(5/8) x = $ 131.25 (Lexi's money) 250 - x = 118.75 (Marie's money) hope this helps~ :)
*October 21, 2011*

**math**

let x = initial number of stickers then set up the equation: x-26 = 2(x-38) x - 26 = 2x - 76 -x = -76 + 26 -x = -50 x = 50 stickers hope this helps~ :)
*October 21, 2011*

**Math**

the equation is not given..
*October 21, 2011*

**chemistry**

recall that the amount of heat absorbed or 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 = (+) when heat is absorbed and (-) when heat is released to get the final temp, we note that Q,...
*October 20, 2011*

**Math**

to do this, we substitute h = 3 to the equation and solve for x: h = -0.005x^2 + x + 3 3 = -0.005x^2 + x + 3 0 = -0.005x^2 + x 0 = x(-0.005x + 1) x = 0 (this is extraneous) x = 1/0.005 = 200 ft hope this helps~ :)
*October 20, 2011*

**Algebra 2**

#1 -16 + 17(5-y²)-(5-y²)² let X = (5-y²) -16 + 17X - X² -(16 - 17X + X²) -(X² - 17X + 16) -(X - 16)(X - 1) substituting back X, -((5-y²) - 16)((5-y²) - 1) or (11 - y²)(4 - y²) #2 [x + y + 1][x² - x(y + 1)+(y + 1)²...
*October 20, 2011*

**chemistry**

recall that the amount of heat absorbed or 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 = (+) when heat is absorbed and (-) when heat is released to get the final temp, we note that Q,...
*October 20, 2011*

**Algebra 2**

you can factor (y-1)² from both, resulting to: (y-1)² [(y-1)² -1] since both terms inside the bracket are perfect squares, we can still factor them, resulting to: (y-1)² [(y-1)-1][(y-1)+1] or (y-1)² [y-2][y] rewriting, y(y-2)(y-1)(y-1) hope this helps~ :)
*October 20, 2011*

**Math**

first represent the unknowns using variables: let 2x = shortest piece let 7x = 2nd longest piece let 11x = longest piece from the second statement, the longest piece is 72 longer than the shortest: 11x = 72 + 2x solving for x, 11x - 2x = 72 9x = 72 x = 8 the length of the ...
*October 15, 2011*

**Algebra**

(a) represent the unknowns using variables: let x = be the smaller odd integer (b) since the difference between any odd integers is always 2, we must add 2 to the smaller odd integer to get the next odd integer. (c) let x + 2 = be the greater odd integer (d) since it is stated...
*October 14, 2011*

**pre-algebra**

just follow PEMDAS: Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction. this is the order in which operation must be performed first. thus, #1. 12^2 + 2[(13+3)] / 2^2] 144 + 2[(16)/4] 144 + 2[4] 144 + 8 152 #2. -4(3)+5[7-6(-4)^2] -12 + 5[7 - 6*(16)] -12 + ...
*October 14, 2011*

**algebra 2**

*i think you mean the ZEROS / ROOTS are 5, 2i and -2i. if so, you can get them by multiplying and expanding, (x-5)(x-2i)(x+2i) note that 1 = sqrt(-1) but an easier method would be, to start with the roots 2i and -2i. recall that q quadratic equation follows the formula, x^2...
*October 14, 2011*

**Math**

(1/8)b = 6 b = 6*8 = 48
*October 13, 2011*

**Algebra**

recall that area of triangle is given by: A = (1/2)*bh where A = area, b = base, and h = height. substituting, 2x^2 + 5x + 3 = (1/2)*(4x+6)*b we do factoring: 2x^2 + 5x + 3 = (1/2)*2*(2x+3)*b (2x+3)*(x+1) = (2x+3)*b then we divide both side by 2x+3. therefore, b = x+1 hope ...
*October 13, 2011*

**algebra**

-18 + |-12| *absolute value: distance from the origin (which is zero). to get this, just take the positive value of the number inside: -18 + 12 -6 hope this helps~ :)
*October 12, 2011*

**Math**

let x = the other number then we set-up the equation. 4*(x-14) = (1/2)*(x+14) 4x - 56 = (1/2)x + 7 4x - 0.5x = 56 + 7 3.5x = 63 x = 18 hope this helps~ :)
*October 9, 2011*

**math**

distributive property
*October 9, 2011*

**Math**

p less than 5: 5 - p p greater than 10: 10 + p 9 times sum of p and 6: 9*(p+6)
*October 8, 2011*

**Math - fractions**

to multiply fractions, just multiply the numerator by numerator, and denominator by denominator. (note: numerator = the number above the fraction bar, while denominator = the number below the fraction bar). thus: 9/16 * 14/15 (9*14) / (16*15) 126 / 240 or 21/40 yd^2 hope this ...
*October 8, 2011*

**Algebra**

for the first problem: -a = 5 to get a, you divide both sides of equation by the numerical coefficient (number before the variable) of a. the numerical coefficient of a is -1, thus: -a/(-1) = 5/(-1) a = -5 for problem 2, 3 and 4: do the same procedure as what we did to problem...
*October 8, 2011*

**Algebra 2**

3y + 6 > 21 3y > 21 - 6 3y > 15 y > 5 graph: h t t p : / / w w w . w o l f r a m a l p h a . com/input/?i=3y+%2B+6+%3E+21
*September 17, 2011*

**Pre-Calc**

you're right. you can do log144/log6 the sign will not be switched.
*September 10, 2011*

**Algebra**

first you represent the unknown with variable: let x = number of hours you rent the canoe then we set-up the equation. since the rate is $10 per hour, we multiply the number of hours (which is x) by 10 and add the usage fee to get the total amount. thus, 10x + 5 = 45 hope this...
*September 10, 2011*

**geometry**

i'm not actually sure about this, but what do you mean "x is the measurement outside the square"? is it the length of the square? because if you have a circle inside a square, you can find right away the area of the whole square using the given radius r. the side...
*September 10, 2011*

**algebra**

if K is between JL, then we can say that the sum of the lengths of JK and KL is equal to JL, or JK + KL = JL 6x + 3x = 27 9x = 27 x = 3 KL = 3x = 9 hope this helps~ :)
*September 10, 2011*

**Chem**

amount of energy absorbed or released is given by Q = mc*(T2 - T1) where Q = heat (in Joules) m = mass (in grams) c = specific heat capacity (J/g-K) T = temperature (in Celsius or Kelvin) *note that Q is (+) if the object absorbed heat, and (-) if it released heat. for water, ...
*September 10, 2011*

**physics 180**

recall that for a projectile motion, the maximum height that the object can reach is given by the formula, h,max = [(vo)^2 * sin^2 (θ)]/(2g) where vo = initial velocity θ = angle of release g = acceleration due to gravity = 9.8 m/s^2 the unknown in the problem is vo...
*September 10, 2011*

**Math**

3(x-2)(x+6) since we want to expand this, we can use the FOIL method: multiply the FIRST terms of each expression inside the parenthesis (x and x), OUTER terms (x and 6), INNER terms (-2 and x) and LAST terms (-2 and 6), and we add the products. therefore, 3*(x*x + x*6 + (-2)*...
*September 10, 2011*

**Algebra 2**

you can rewrite -27 as -26 16/16 (since 16/16 is just equal to 1). thus, -26 16/16 + 1/16 -26 + (16-1)/16 -26 15/16
*September 10, 2011*

**Algebra 2**

x^3 + y^(-2) to evaluate means you substitute the value of x and y which are given in the problem. first, recall some laws of exponents. if a term is raised by a negative number, it is also equal to its reciprocal raised to the that number but positive in sign. for example, 5...
*September 10, 2011*

**maths**

first we represent unknowns using variables. let x = number of girls in the hall since the the total number of children is 335, let 335 - x = number of boys in the hall then we set-up the equation. since 1/4 of the girls and 2/5 of the boys are swimmers and their total is 110...
*September 10, 2011*

**Algebra 2**

recall some laws of exponents. a term which is raised by a negative number is also equal to its reciprocal raised to that number with positive sign. for example, 3^(-2) = (1/3)^2 thus, we can rewrite it as, (1/x)^4 substituting x = 2, (1/2)^4 (1/2)*(1/2)*(1/2)*(1/2) 1/16 hope ...
*September 10, 2011*