Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - [SOLVED] Help! Many algebra problems :(

  1. #1
    Elite00
    Guest

    [SOLVED] Help! Many algebra problems :(

    Hey, I'm a high school student and recently came back to school after a long period of absence (appendicitis ) and my teacher gave me a 2-page packet with a bunch of problems from the chapters I missed, and he gave me all the notes to learn how to do them on my own. I am struggling very much and these problems are due tomorrow! I would greatly appreciate if anyone could help me out, even if you just solve 1 or 2.

    1. Find the slope of the line between the points (-10, -4) and (-3, -3)

    2. Write the equation in slope-intercept form for the line that contains the point (-2,7) and is parallel to y = 3x 4

    3. Write an equation in slope-intercept form for the line that contains the point (8,-1) and is perpendicular to the graph y=4x-3

    4. Solve for x in the proportion 6x-3/9 = 3x/8

    5. Use the table provided to (a) Write the linear regression equation and (b) tell how many students were enrolled in AP statistics for the 4th year.
    Year 1 2 3 5 6 7
    Enrollment 33 30 43 47 58 65

    6. Solve for x in the following equation 2/5x + 6/5 = x-3

    7. Solve and graph the solution(s) for the following inequalities 2x 3 > 1 or 3x+7 < 1

    8. Solve and graph the following |3x+2| > 4

    9. Simplify the expressions completely by leaving only positive exponents:
    (3x^3 y^2 / y^-1)^3 (x^-2 / 2y^-1) ^-3(3y/x^-4)^-2

    10. Tell whether or not the following equation represents a function. Then give the domain, range, and zeros. f(x) = 9 x^2

    11. Let f(x) = 4x^2 and g(x) = 2x-3. Find each of the following:

    a. (f g)(x) b. (f x g)(x) c. g ( f (x) )

    12. Find the inverse of the following function. Then use the composition of functions to prove that your equations are inverses of each other. f ( x ) = x-1 / 3

    13. Graph the following piecewise function and give the domain, range, and zeros.
    f (x ) = {2x + 3, x<4
    {x -1, 4 < x < 9

    14. Solve the following systems of equations:
    a. {s = 9 + 5t b. {3x + 2y =12
    {2s = 30 2t {5x-3y = -18

    15. A lab technician is trying to make an 8% solution by combining a 2% and 12% solution. If he is trying to make a 150mL solution, how many mL of each solution should be mixed?

    16. Graph, shade the feasible region and label the corner points for the following system of inequalities: {y < -2x -2
    {y > 1/2x
    {x > -3

    17. Write the pair of parametric equations as a single equation of and y
    {x ( t ) = 3t -1
    {y ( t ) = t + 4

    18. Let A = [ 8 4 ] and b= [6 -2] Show each of the following: A B and BA
    [4 1 ] [0 4]

    19. Let C = [2 4] Find the inverse of C.
    [1 0]

    20. Solve the following system of equations: { -3x + 6y 4z = 8
    {x + 2z 4y = -3
    {8y=z

    21. Solve 4(x+5)^2 = 20. express the answer exactly and if necessary, rounded to the nearest tenth.

    22. Factor 2x^2 10x 48 =0 and solve for x

    23. Find the exact values of x in the following quadratic equation. 2x^2 5x -12 = 0

    24. Simplify the following complex number: 3 + i / 2 3i

    25.Solve the Inequality 4x 1 > 8 x^2 and graph it on a number line.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    1. Find the slope of the line between the points (-10, -4) and (-3, -3)
    The slope between two distinct points (x_1, y_1) and (x_2, y_2) is defined as:
    m = \frac{y_2 - y_1}{x_2 - x_1}.

    So
    m = \frac{-3  - -4}{-3 - -10} = \frac{-3 + 4}{-3 + 10} = \frac{1}{7}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    2. Write the equation in slope-intercept form for the line that contains the point (-2,7) and is parallel to y = 3x 4
    The slope-intercept form of a line is y = mx + b where m is the slope and the point (0, b) is the y-intercept: the point where the line crosses the y-axis.

    So the slope of the line parallel to y = 3x - 4 is m = 3. (Parallel lines have the same slope.) We know this line passes through the point (-2, 7), so applying the slope-intercept form for the line:
    7 = 3 \cdot -2 + b

    7 = -6 + b

    b = 13

    So the line will be y = 3x + 13.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    3. Write an equation in slope-intercept form for the line that contains the point (8,-1) and is perpendicular to the graph y=4x-3
    This is done in the same way as problem 2, except that the relationship between the slopes of perpendicular lines is
    m_2 = \frac{-1}{m_1}

    So the slope of the line you are interested in is m =  -\frac{1}{4}.

    -Dan
    Last edited by topsquark; November 27th 2006 at 01:12 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by topsquark View Post
    This is done in the same way as problem 2, except that the relationship between the slopes of parallel lines is
    m_2 = \frac{-1}{m_1}

    So the slope of the line you are interested in is m =  -\frac{1}{4}.

    -Dan
    just to clarify, he means perpendicular
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Quick View Post
    just to clarify, he means perpendicular
    I've fixed it in the original post. Thanks for the spot, Quick.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    4. Solve for x in the proportion 6x-3/9 = 3x/8
    I presume this is
    \frac{6x - 3}{9} = \frac{3x}{8}

    \frac{3(2x - 1)}{9} = \frac{3x}{8}

    \frac{2x - 1}{3} = \frac{3x}{8}

    Now, I'm going to "clear the fractions." The least common multiple of 3 and 8 is 24:

    24 \cdot \frac{2x - 1}{3} = 24 \cdot \frac{3x}{8}

    8 (2x - 1) = 3 (3x)

    16x - 8 = 9x

    16x - 9x = 8

    7x = 8

    x = \frac{8}{7}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    6. Solve for x in the following equation 2/5x + 6/5 = x-3
    I presume this is:
    \frac{2}{5x} + \frac{6}{5} = x - 3

    I'm going to add the fractions: The least common multiple of 5x and 5 is 5x.

    So
    \frac{2}{5x} + \frac{6}{5} \cdot \frac{x}{x} = x - 3

    \frac{2 + 6x}{5x} = x - 3

    5x \cdot \frac{2 + 6x}{5x} = 5x (x - 3)

    2 + 6x = 5x^2 - 15x

    5x^2 - 21x - 2 = 0

    As this doesn't factor, use the quadratic formula:
    If ax^2 + bx + c = 0 then x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    So:
    x = \frac{21 \pm \sqrt{21^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5}

    x = \frac{21 \pm \sqrt{481}}{10}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    7. Solve and graph the solution(s) for the following inequalities 2x 3 > 1 or 3x+7 < 1
    First:
    2x - 3 > 1

    2x > 4

    x > 2

    The solution set here is (2, \infty).

    Second:
    3x+7 < 1

    3x < -6

    x < -2

    The solution set is (-\infty, -2)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    8. Solve and graph the following |3x+2| > 4
    This is actually two inequalities:
    3x + 2 > 4
    and
    3x + 2 < -4

    The first one gives:
    3x + 2 > 4

    3x > 2

    x > \frac{2}{3}

    and the second one gives:
    3x + 2 < -4

    3x < -6

    x < -2

    Combining both of these gives a solution set of:
    (-\infty, -2) \cup (\frac{2}{3}, \infty)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    9. Simplify the expressions completely by leaving only positive exponents:
    (3x^3 y^2 / y^-1)^3 (x^-2 / 2y^-1) ^-3(3y/x^-4)^-2
    \left ( \frac{3x^3y^2}{y^{-1}} \right )^3 \left ( \frac{x^{-2}}{2y^{-1}} \right )^{-3} \left ( \frac{3y}{x^{-4}} \right )^{-2}

    The rules of exponents are as follows: (Let's keep it simple and assume for now that a, b, x, y are positive integers.)
    a^x a^y = a^{x+y}
    \frac{a^x}{a^y} = a^{x-y}
    (ab)^x = a^x b^x
    a^0 = 1
    a^{-x} = \frac{1}{a^x}
    (a^x)^y = a^{xy}

    The first thing to do is get rid of those pesky negative exponents:
    (3x^3y^2y)^3 \left ( \frac{y}{2x^2} \right )^{-3} (3x^4y)^{-2}

    (3x^3y^3)^3 \left ( \frac{2x^2}{y} \right )^{3} \left ( \frac{1}{3x^4y} \right )^{2}

    (3^3x^{3 \cdot 3}y^{3 \cdot 3}) \left ( \frac{2^3x^{2 \cdot 3}}{y^3} \right ) \left ( \frac{1}{3^2x^{4 \cdot 2}y^2} \right )

    (27x^9y^9) \left ( \frac{8x^6}{y^3} \right ) \left ( \frac{1}{9x^8y^2} \right )

    \frac{(27x^9y^9)(8x^6)}{(y^3)(9x^8y^2)}

    \frac{27 \cdot 8}{9} \cdot \frac{x^9 \cdot x^6}{x^8} \cdot \frac{y^9}{y^3 \cdot y^2}

    (3 \cdot 8) \cdot \frac{x^{15}}{x^8} \cdot \frac{y^9}{y^5}

    24x^7y^4

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    10. Tell whether or not the following equation represents a function. Then give the domain, range, and zeros. f(x) = 9 x^2
    A function is a relation (a set of ordered pairs (x,f(x)) ) such that there is only one f(x) for each x.

    A simple way to test if you have a function is to graph f(x). If you can draw ANY vertical line through the graph that cuts the graph in more than one place then f(x) is not a function.

    The graph of f(x) = 9 - x^2 shows that it is indeed a function. (See attachment below.)

    There are no values of x such that f(x) is not defined. (Typically such a value of x would make the denominator of the function 0. We don't have any fractions here.) So the domain of the function is all real numbers: (-\infty, \infty).

    The range is a bit trickier. Note that the function has a maximum value: f(max) = 9. There is no lower bound to the function, so the range of the function is (-\infty, 9] .

    Zeros of a function are values of x such that f(x) = 0. So we need to solve
    f(x) = 0 = 9 - x^2

    x^2 = 9

    x = \pm 3

    So the zeros of the function are x = -3 and x = +3.

    -Dan
    Attached Thumbnails Attached Thumbnails [SOLVED] Help! Many algebra problems :(-function.jpg  
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    11. Let f(x) = 4x^2 and g(x) = 2x-3. Find each of the following:

    a. (f g)(x) b. (f x g)(x) c. g ( f (x) )
    f(x) = 4x^2
    g(x) = 2x - 3

    a) (f - g)(x) = f(x) - g(x) = (4x^2) - (2x - 3) = 4x^2 - 2x + 3

    b) (f \times g)(x) = f(x) \times g(x) = (4x^2) \cdot (2x - 3) = 8x^3 - 12x^2

    c) g(f(x)) = g(4x^2) = 2(4x^2) - 3 = 8x^2 - 3

    -Dan
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    12. Find the inverse of the following function. Then use the composition of functions to prove that your equations are inverses of each other. f ( x ) = x-1 / 3
    I'm assuming this is: f(x) = \frac{x - 1}{3}

    A function and its inverse have a kind of symmetry: calling y = f(x), the graph of f(x) and it's inverse are reflections over the line y = x. This means that, in order to find the inverse of the function y = f(x) what we need to do is switch the x and y in the equation: x = f(y). Once we solve this for y we have our inverse function.

    So:
    y =  \frac{x - 1}{3}

    Switch x and y:
    x =  \frac{y - 1}{3}

    Now solve for y:
    x =  \frac{y - 1}{3}

    3x = y - 1

    y = 3x + 1

    So the inverse function is g(x) = 3x + 1. (Typically we use the notation f^{-1}(x) for the inverse function.)

    Now we need to show that these functions are inverses of each other. If this is true then we should be able to show that
    f(g(x)) = x
    and
    g(f(x)) = x

    I'll do the first one.
    f(g(x)) = f(3x + 1) = \frac{(3x + 1) - 1}{3} = \frac{3x}{3} = x (Check!)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Elite00 View Post
    14. Solve the following systems of equations:
    a. {s = 9 + 5t b. {3x + 2y =12
    {2s = 30 2t {5x-3y = -18
    a)
    s = 9 + 5t
    2s = 30 - 2t

    There are several ways to solve systems of equations. This one is called the "substitution method." It's occasionally more trouble that the other methods, but it has the advantage of being one of the more systematic methods.

    What we want to do is solve one of the equations for one of the unknown variables. We then insert this value into the second equation. This provides us with an equation in a single unknown. We can solve this and use this value in either of the original equations to find the value of the other variable.

    For example:
    s = 9 + 5t
    2s = 30 - 2t

    As it happens, the top equation is already solved for s. I will use this value of s in the second equation:
    2(9 + 5t) = 30 - 2t

    This is an equation in a single unknown, t, which we can solve:
    18 + 10t = 30 - 2t

    12t = 12

    t = 1

    We now use this value of t in either of the original equations to get a value for s:
    s = 9 + 5 \cdot 1 = 14

    So the solution to the set of equations is (s, t) = (14, 1).

    You can do the same for problem b). I get that (x, y) = (0, 6).

    -Dan
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Algebra 2 problems.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 29th 2009, 10:50 AM
  2. Algebra, Problems For Fun (18)
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: June 18th 2009, 03:22 PM
  3. Algebra, Problems For Fun (19)
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: June 18th 2009, 01:20 PM
  4. [SOLVED] Digital clock algebra problems
    Posted in the Algebra Forum
    Replies: 0
    Last Post: March 29th 2007, 07:53 AM
  5. [SOLVED] algebra word problems
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 30th 2006, 01:20 PM

Search Tags


/mathhelpforum @mathhelpforum