Monday

January 23, 2017
Total # Posts: 1,998

**algebra 1**

(z - 4)/(z + 7) = 1/5 cross-multiplying, 5(z - 4) = z + 7 5z - 20 = z + 7 4z = 27 z = 27/4

*May 23, 2012*

**Geometry**

tan A * tan B = 1

*May 23, 2012*

**math**

sorry, i could not understand #15. #16. you have to plot this. But if you already know some properties of circle (in terms of its equation), you can skip plotting. The equation is a graph of circle, with center at (-6,-4) and radius = 6. The center is the position of source, ...

*May 23, 2012*

**circles**

At first, I was having difficulty understanding the given. I thought the PQ, PR and QR are the lengths of the triangle, but they are actually the arc (angle measure). Anyway, to get x, we know that the measure of a circle is 360 degrees. We get the sum of arcs PQ, PR and QR ...

*May 23, 2012*

**math**

3.62 x 10^7

*May 23, 2012*

**circles and arcs**

At first, I was having difficulty understanding the given. I thought the PQ, PR and QR are the lengths of the triangle, but they are actually the arc (angle measure). Anyway, to get x, we know that the measure of a circle is 360 degrees. We get the sum of arcs PQ, PR and QR ...

*May 23, 2012*

**chemistry**

recall that one mole of any substance is equal to Avogadro's number, which is 6.02 x 10^23 this number can be atoms, molecules, ions, etc. to get the number of moles, we just divide the given by the Avogadro's number: 2.8 x 10^20 / 6.02 x 10^23 = ?

*May 23, 2012*

**algebra**

[-16(x^7)(y^4)z]/[20(x^2)(z^3)] -4(x^5)(y^4)/(5z^2)

*May 23, 2012*

**PHYSICAL SCIENCES**

A. Argon all noble gases, except Helium, has eight valence electrons.

*May 23, 2012*

**MATHS**

that is true, if 'ln' is used, not 'log'. Well, in some books, ln and log seem interchangeable -they really mean ln, not log. If we rewrite the expression using ln, not log, e^(x + 3lnx) we can separate this into: (e^x)[e^(3lnx)] note that we can further ...

*May 23, 2012*

**math**

*sorry i mean x = 3/4

*May 23, 2012*

**math**

6 = 4x + 3 6 - 3 = 4x 3 = 4x x = 4/3

*May 23, 2012*

**maths**

work backwards: 2(2(2(2(0+1)+1)+1)+1) = 30 liters

*May 22, 2012*

**Math**

525

*May 22, 2012*

**Math**

let x = age of Travis let x-4 = age of Ellie (since Travis is four years older than Ellie) let 2(x-4) = age of Mac (since he is twice older than Ellie) since the sum of their ages is 44, x + x - 4 + 2(x - 4) = 44 2x - 4 + 2x - 8 = 44 4x - 12 = 44 4x = 56 x = 14 yrs old (Travis...

*May 22, 2012*

**Math**

recall that volume of cone is given by V = (1/3)*pi*(r^2)*h where r = radius which is half of diameter h = height and pi = constant equal to 3.14 substituting, V = (1/3)*3.14*[(5/2)^2]*7 now solve for V.

*May 22, 2012*

**math algebra**

let x = first number let x+1 = second number let x+2 = third number let x+3 = fourth number *note that there is +1 for every number since they are consecutive. therefore, x + x+1 + x+2 + x+3 = -2 4x + 6 = -2 4x = -8 x = -2 x+1 = -1 x+2 = 0 x+3 = 1 hope this helps~ :)

*May 21, 2012*

**math trig**

ln(2x-5) = 4 we raise both sides by e: e^(ln(2x-5)) = e^4 the e and ln will be cancelled, leaving only the 2x - 5: 2x - 5 = e^4 2x = e^4 + 5 x = (e^4 + 5)/2 hope this helps~ :)

*May 21, 2012*

**math**

Oh sorry. I thought it's outside. Nope, it can't be moved outside. 2^(x+1) = 9 2^(x+1) = 3^2 then we take the log of both sides: (x+1)*(log 2) = 2*(log 3) x + 1 = 2(log 3) / (log 2) x = [2(log 3) / (log 2)] - 1 or we can also rewrite the 1 as (log 2)/(log 2), so x = [2...

*May 21, 2012*

**math**

2^x + 1 = 9 2^x = 9 - 1 2^x = 8 then we rewrite 8 as power of 2: 2^x = 2^3 finally, since they have the same base (which is 2) we equate their exponents: x = 3 hope this helps~ :)

*May 21, 2012*

**math**

No. First to multiply fractions, we change mixed number into improper fraction. The mixed number here is 1 1/4. To change this, we multiply the denominator (the 4) to the constant (the first 1, not the 1 above the bar line) and add this to numerator. thus, 1 1/4 = (1*4 + 1)/4...

*May 21, 2012*

**maths**

recall these properties: -all sides of rhombus are equal. -the diagonals of rhombus are perpendicular (forms four 90 degree angles) -area of rhombus, if diagonals are given, can be calculated by A = d1 * d2 /2 where d1 & d2 are the diagonal lengths -side of a rhombus, if ...

*May 21, 2012*

**maths**

I think it has a typo. (what is "1nd")

*May 21, 2012*

**algebra**

for this problem, we will substitute the given to the equation p = mx + b, and then we solve simultaneously (two equations, two unknowns): 25 = 56m + b 20 = 66m + b to solve this, we can multiply one equation by -1 so that if we add the equations, the b will be cancelled: -25...

*May 21, 2012*

**maths**

let x = first number let 8-x = second number then we set up the equation. It says that the sum of their squares must be 34. thus, x^2 + (8-x)^2 = 34 then we solve for x: x^2 + 64 - 16x + x^2 = 34 2x^2 - 16x + 64 = 34 2x^2 - 16x + 30 = 0 x^2 - 8x + 15 = 0 since it's ...

*May 21, 2012*

**maths**

recall that an isosceles triangle has two equal sides. thus, to find this length, recall that the perimeter of a triangle is given by P = a + b + c but since two sides are equal, we can rewrite this as P = 2a + c substituting, 32 = 2a + 12 20 = 2a a = 10 now to find the area, ...

*May 21, 2012*

**Geometry**

recall that the volume of an oblique cone is given by V = (1/3)*(πr^2)*h where r = radius of circle (the base) h = height since diameter is equal to the height, we can say that h = d = 2r substituting, V = (1/3)π(r^2)(2r) 18π = (2/3)π(r^3) 18*(3/2) = r^3 27...

*May 21, 2012*

**Algebra (re: %)**

60/70 * 100 = 85.7%

*May 21, 2012*

**Trigonometry**

To combine the two log terms, first, the terms must have the same base (the number subscript of log). In this case, if there is no number subscript of log, the base is equal to 10. The two terms have the same base, 10. Now, recall that to combine log terms of the same base, *...

*May 19, 2012*

**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*