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Math Help - Please Help Me Plzz I Have General Maths Exam Tomorrow And I Dont Kow What I Am Doing

  1. #1
    duygu
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    Please Help Me Plzz I Have General Maths Exam Tomorrow And I Dont Kow What I Am Doing

    FRIENDS PLEASE HELP ME SOLVE THESE BECASE IM ALLOWED TO TAK IN A 2 PAGE CHEAT SHEET INTO THE EXAM AND THESE QUESTIOSN AREON MY EXAM REVISIONS SHEET. IF I HAVE THE FORMULAS FOR THESE QUESTION THEN WHEN I AM IN THE EXAM I CNA LOOK ONTO THIS SHEET AND FIGURE OTU THE REST OF THE QUESTIONS. PLEASE FRIENDS HELP ME. IF I FAIL THIS EXAM I FAIL UNIT 2 OF GENERAL MATHS THEN I HAVE TO DO AN EXTRA UNIT FOR VCE NEXT YEAR AN DI DONT WANT TO DO THAT. I DOTN UDERSTAND GENERAL MATHS THAT WELL BECAUSE HALF THE CLASSES WE HAD THIS UNIT OUR TEACHER WAS AWAY AND WE COULDNT DO ANY PROPER MATHS WORK. MY EXAM IS TOMORROW AND I NEED HELP. THANKS GUYS.

    the gradient of the line which passes through the points (3, y) and (2, 5) is
    -3. which is the value of y?
    A. -3
    B. -3
    C. -1
    D. 1
    E. 2

    which pair of lines are parralel?
    A. y=3x+2 and y=2x+3
    B. y=1.5x-3 and 6y-3x=6
    C. 3x-4y=12 and 4x-3y=12
    D. y=2x and y=2+3x
    E. y=3-2x and y=-2

    to sketch the graph of y=2-3x using the gradient-intercept method, we need to move, beginning from the y-intercept:
    A. 3 units up and 2 units right
    B. 3 units up ad 2 units left
    C. 3 units down and 1 unit right
    D. 3 units down and 1 unit left
    E. 3 units down only

    a taxi meter is adjusted to charge fars (f) for the dstance travelled (d) in kilometres according to the equation F=1.5d+3.50 the distance covered on a fare of $14 is:
    A. 5km
    B. 7km
    C. 20.5km
    D. 24km
    E. none of these

    which of the following is the same as 3(2x-1)=4x+5
    A. 2x=8
    B. 6x=4x+5
    C. 6x=4x=5
    D. 6x=5x+3
    E. 6x=4x+3

    which of the following satisfies the equation (1-3x/4)+(x/5)=2
    A. -4
    B. -2
    C. 2
    D. 3
    E. none of these

    he perimeter of a regular hexagon is 15cm more tha the perimeter of a square with the same side length. the length of the side of a hexagon is:
    A. 7.5cm
    B. 3.75cm
    C. 2.5cm
    D. 4cm
    E. 15cm

    if P=(6P-5/6) is transposed to make R the subject, then:
    A. R=(6P-5/2)
    B. R=3P-5
    C. R=3P-(5/2)
    D. R=6P-(5/2)
    E. R=2(6P-5)

    if x=-3, then the value of y in the relation 2x-3y=6 is:
    A. -1
    B. -2
    C. -3
    D. -4
    E. 4

    the general form of gradient-intercept equation is y=mx+c. if y=4, x=3, c=-2 then m is equal to:
    A. -2
    B. 2
    C. 3
    D. -3
    E. 4

    the sum of solutions to the simultaneous equations y-x=12 and 2y+x=6 is:
    A. 0
    B. 12
    C. 6
    D. 36
    E. -6

    the sum of first 5 terms of geometric sequence with the first term 2 and common ratio -2 is:
    A. 22
    B. 23
    C. 24
    D. 25
    E. none of these

    write down the domain and the range of the relation {(2, 3), (2, 4), (5, 7), (-1, 8)}.

    rearange the equation 2x-y=5 to make y the subject.

    find the gradient of the line joining the points (-2, -6) and (3, 4).

    fidn the equation of a line which passes through the points (3, 5) and (5, 11)

    rewrite the equation 4x-2y=6 in the form y=mx+c and state its gradient and the y-intercept.

    the angle of elevation from a melbourne airport runway to an approching plane at an altitude of 300m is 7 degrees. how far is the plane from the runway?

    a cross country runner starts at checkpoint A and runs 8km on a bearing of 50degreesT to reach checkpoint B then heads directly east for 10km to checkpoint C. how far is checkpoint C from the starting point A?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by duygu View Post
    FRIENDS PLEASE HELP ME SOLVE THESE BECASE IM ALLOWED TO TAK IN A 2 PAGE CHEAT SHEET INTO THE EXAM AND THESE QUESTIOSN AREON MY EXAM REVISIONS SHEET. IF I HAVE THE FORMULAS FOR THESE QUESTION THEN WHEN I AM IN THE EXAM I CNA LOOK ONTO THIS SHEET AND FIGURE OTU THE REST OF THE QUESTIONS. PLEASE FRIENDS HELP ME. IF I FAIL THIS EXAM I FAIL UNIT 2 OF GENERAL MATHS THEN I HAVE TO DO AN EXTRA UNIT FOR VCE NEXT YEAR AN DI DONT WANT TO DO THAT. I DOTN UDERSTAND GENERAL MATHS THAT WELL BECAUSE HALF THE CLASSES WE HAD THIS UNIT OUR TEACHER WAS AWAY AND WE COULDNT DO ANY PROPER MATHS WORK. MY EXAM IS TOMORROW AND I NEED HELP. THANKS GUYS.
    First off, don't yell.

    Second, you knew after the first exam that you needed to pass the second one, or fail out. You had plenty of chances to seek out help during that time interval. And you are coming here the day before the exam begging for answers to material that if you don't know how to do now, you are unlikely to learn for the next test.

    This strategy is not helping you with your chances. Next time, try to plan ahead better.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
    Joined
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    Quote Originally Posted by duygu View Post
    the gradient of the line which passes through the points (3, y) and (2, 5) is
    -3. which is the value of y?
    A. -3
    B. -3
    C. -1
    D. 1
    E. 2
    The slope between two points (x_1, y_1) and (x_2, y_2) is defined to be
    m = \frac{y_2 - y_1}{x_2 - x_1}

    Quote Originally Posted by duygu View Post
    which pair of lines are parralel?
    A. y=3x+2 and y=2x+3
    B. y=1.5x-3 and 6y-3x=6
    C. 3x-4y=12 and 4x-3y=12
    D. y=2x and y=2+3x
    E. y=3-2x and y=-2
    Two lines are parallel if their slopes are equal. So what are the slopes of these lines?

    Quote Originally Posted by duygu View Post
    to sketch the graph of y=2-3x using the gradient-intercept method, we need to move, beginning from the y-intercept:
    A. 3 units up and 2 units right
    B. 3 units up ad 2 units left
    C. 3 units down and 1 unit right
    D. 3 units down and 1 unit left
    E. 3 units down only
    Remember, slope is defined as
    m = \frac{y_2 - y_1}{x_2 - x_1}

    For a given change in the x value ( \Delta x) we can calculate the change in the y value ( \Delta y) as
    \Delta y = m \Delta x

    So for example, say we have the line y = -2x + 3. We plot the intercept point (0, 3). Now we want the point to the right: x = 1, so \Delta x = 1. Thus \Delta y = -2(1) = -2. Thus the y value for the point on the line with x = 1 is going to be 3 + -2 = 1.

    See if you can apply this to your problem.

    Quote Originally Posted by duygu View Post
    a taxi meter is adjusted to charge fars (f) for the dstance travelled (d) in kilometres according to the equation F=1.5d+3.50 the distance covered on a fare of $14 is:
    A. 5km
    B. 7km
    C. 20.5km
    D. 24km
    E. none of these
    Solve
    14 = 1.5d + 3.50

    14 - 3.50 = 1.5d + 3.50 - 3.50

    10.5 = 1.5d

    You do the rest.

    Quote Originally Posted by duygu View Post
    which of the following is the same as 3(2x-1)=4x+5
    A. 2x=8
    B. 6x=4x+5
    C. 6x=4x=5
    D. 6x=5x+3
    E. 6x=4x+3
    Expand out the left hand side and start reducing the equation to a solution until you get one of the forms listed in the answer.
    3(2x-1)=4x+5

    3 \cdot 2x + 3 \cdot (-1) = 4x + 5

    6x - 3 = 4x + 5

    6x - 3 + 3 = 4x + 5 + 3

    6x = 4x + 8

    You take it from here.

    Quote Originally Posted by duygu View Post
    which of the following satisfies the equation (1-3x/4)+(x/5)=2
    A. -4
    B. -2
    C. 2
    D. 3
    E. none of these
    Good heavens! If you can't solve it, just plug in each possible answer to see which one is right.

    Quote Originally Posted by duygu View Post
    he perimeter of a regular hexagon is 15cm more tha the perimeter of a square with the same side length. the length of the side of a hexagon is:
    A. 7.5cm
    B. 3.75cm
    C. 2.5cm
    D. 4cm
    E. 15cm
    Take the side length to be s. Then we know that
    6s = 4s + 15

    Solve for s.

    Quote Originally Posted by duygu View Post
    if P=(6P-5/6) is transposed to make R the subject, then:
    A. R=(6P-5/2)
    B. R=3P-5
    C. R=3P-(5/2)
    D. R=6P-(5/2)
    E. R=2(6P-5)
    There is no R in this equation.

    Quote Originally Posted by duygu View Post
    if x=-3, then the value of y in the relation 2x-3y=6 is:
    A. -1
    B. -2
    C. -3
    D. -4
    E. 4
    Plug in x = -3:
    2(-3) - 3y = 6

    -6 - 3y = 6

    Solve for y.

    Quote Originally Posted by duygu View Post
    the general form of gradient-intercept equation is y=mx+c. if y=4, x=3, c=-2 then m is equal to:
    A. -2
    B. 2
    C. 3
    D. -3
    E. 4
    Solve
    4 = 3m - 2
    for m.

    Quote Originally Posted by duygu View Post
    the sum of solutions to the simultaneous equations y-x=12 and 2y+x=6 is:
    A. 0
    B. 12
    C. 6
    D. 36
    E. -6
    -x + y = 12
    x + 2y = 6

    Add the two equations:
    (-x + x) + (y + 2y) = (12 + 6)

    3y = 18

    Solve for y.

    Put this value of y into either of the two original equations, then solve for x.

    Then you are looking for the value of x + y.

    Quote Originally Posted by duygu View Post
    the sum of first 5 terms of geometric sequence with the first term 2 and common ratio -2 is:
    A. 22
    B. 23
    C. 24
    D. 25
    E. none of these
    Wow. This material jumps around a rather lot for a single test.

    The general geometric series can be written as
    a_{n + 1} = a_nr^{n - 1}
    where a_1 is the first term and r is the geometric ratio (ie. your "common ratio.")

    The sum of the first n terms will be:
    S_n = \frac{a_1 (1 - r^n)}{1 - r}

    Quote Originally Posted by duygu View Post
    write down the domain and the range of the relation {(2, 3), (2, 4), (5, 7), (-1, 8)}.
    This is just a definition. Look it up in your textbook.

    Quote Originally Posted by duygu View Post
    the angle of elevation from a melbourne airport runway to an approching plane at an altitude of 300m is 7 degrees. how far is the plane from the runway?
    Sketch the triangle. The altitude of the plane is the vertical distance between the ground and the plane. The horizontal leg of the triangle is the horizontal distance the plane is from the runway to the observer. The straightline distance between the observer and the plane is what you are trying to find. The angle 7 degrees is the angle between the horizontal leg of the triangle and the sloped line between the observer and the plane.

    You can find your answer by using the definition of the sine of an angle.

    -Dan
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