my math teacher gave us a 100 question packet to do for monday, but i have to write a speech, and also have an essay due in english on frankenstein... ive got 70ish of the questions done, can someone help me with the rest of the questions, theyre not hard, but it wuld really help me out. also, could you show me how you worked them out, this is a review.
if you need the questions to be bigger, theyre bmp files (you can save them and zoom).
page one
page two
here is the second set of questions. you said you can do the number line part so i won't bother with that. i will just simplify for you.
now we treat inequality signs just like equal signs, except in two cases. we turn the sign the other way if:
1) we multiply through by a negative number
2) we take the inverse of both sides
otherwise, we solve for x as if we were working with a regular equation.
1) 4 - 3x <= 8 - x
=> -2x <= 4 ...............i added x to both sides and then -4 to both sides
=> x >= -2 ................i multiplied through by -2, so i turned the sign around
2) |2x + 5| - 4 <= 3
=> |2x + 5| <= 7 ...........added 4 to both sides
now we have an absolute value inequality. absolute values give positive output even if what's inside the absolute value signs is negative. so we have to account for the posibility that the 2x + 5 on the inside is negative, and the posibility that it is positive. to do this, we split it into two inequalities.
=> 2x + 5 <= 7...............or.............. -(2x + 5) <= 7
=> 2x <= 2....................or..............2x + 5 >= -7
=> x <=1.......................or..............2x >= -12
.................................................. ...x >= -6
3) 5 - 3x <= -1...............or..............5 - 2x > 7
deal with each separately. when plotting them on the number line, make sure both solutions do not conflict with each other
5 - 3x <= -1...............or..............5 - 2x > 7
=> -3x <= -6..............or..............-2x > 2
=> x >= 2...................or..............x < -1
4) x - 2x - 3 > 0
=> -x - 3 > 0
=> -x > 3
=> x < -3
here's the last question on page one.
note that:
- two lines are parallel if they have the same slope
- two lines are perpendicular if their slopes are negative inverses of each other
- the slope-intercept form for the equation of a line is written in the form:
y = mx + b
when in this form, the slope is m and the y-intercept is b
now on to your problem.
x + 2y = 6 .............let's rewrite this in the slope-intercept form.
=> 2y = -x + 6
=> y = (-1/2)x + 3
so we see that the slope of our line is -1/2
this means a line parallel to it must have a slope of -1/2 as well
we want a line with slope -1/2 and y-intercept -1
y = mx + b
when m = -1/2 and b = -1, we have
y = (-1/2)x - 1 ............this is the slope-intercept form of the line we want, let's change it to the standard form
=> 2y = -x - 2
=> x + 2y = -2
so by now you must have realized something. the equation is the same, except for what's on the right. do you see what the connection is?
page 2, question 3
2x - 3y - 6 = 0
3y - 8 = 6x - 2
=> 2x - 3y = 6
=> -6x + 3y = 6
now these are simultaneous equations, so we have:
2x - 3y = 6 ...................(1)
-6x + 3y = 6 .................(2)
=> -4x = 12
=> x = -3
but 2x - 3y = 6
=> 2(-3) - 3y = 6
=> -6 - 3y = 6
=> -3y = 12
=> y = -4
Second set of questions, page 2
1) x + 3 = x - 3
there is no solution to this, check to make sure it's not a typo
2) 3/(x - 2) + 4/x = 1 ...........combine the fractions on the left
=> (3x + 4(x - 2))/[x(x - 2)] = 1
=> (7x - 8)/[x(x - 2)] = 1 .............now multiply both sides by x(x - 2)
=> 7x - 8 = x(x - 2)
=> 7x - 8 = x^2 - 2x
=> x^2 - 9x + 8 = 0
=> (x - 8)(x - 1) = 0
=> x = 8 or x = 1
3) (x + 1)(2x - 3) = x + 2 ..........expand the left
=> x^2 - x - 3 = x + 2
=> x^2 - 2x - 5 = 0
=> By the quadratic formula:
x = [2 +/- sqrt(4 + 20)]/2
x = (2 + sqrt(24))/2 or x = (2 - sqrt(24))/2
4) x^3 - 8 = 0
=> x^3 - 2^3 = 0 ..............this is the difference of two cubes
=> (x - 2)(x^2 + 2x + 4) = 0
=> x = 2
or
x = [-2 +/- sqrt(4 - 16)]/2
x = [-2 +/- sqrt(-12)]/2
x = [-2 +/- 2sqrt(-3)]/2
x = -1 + sqrt(3)*i or x = -1 - sqrt(3)*i
Third set of questions, page 2
f(x) = x^2 + 4x - 5
for the x-intercept, set y = 0
=> x^2 + 4x - 5 = 0
=> (x + 5)(x - 1) = 0
=> x + 5 = 0 or x - 1 = 0
=> x = -5 and x = 1 are the x-intercepts
for the y-intercept, set x = 0
=> y = 0^2 + 4(0) - 5
=> y = -5 is the y-intercept
For a quadratic of the form: y = ax^2 + bx + c, the vertex is given by:
x = -b/2a
so the vertex for this parabola is given by:
x = -4/2 = -2
when x = -2
y = (-2)^2 + 4(-2) - 5
=> y = 4 - 8 - 5
=> y = -9
so the vertex is (-2,-9)
the domain of a function is the set of all inputs (in this case, x-values) for which the function is defined.
f(x) = x^2 + 4x - 5 is a polynomial, therefore, it is defined for all x-values and so the domian is all real x.
so dom(f) = all real x or in interval form: (-infinity, infinity)
the range of a function is the set of all outputs (in this case, the set of y-values for which the function is defined).
this is an upward opening parabola, therefore it has a minimum value, and so, all y's below this value are NOT in the range. the minimum value occurs at the vertex, so the range is given by:
ran(f) = all y greater than or equal to -9 or in interval form: [-9, infinity)
in this case, the axis of symmetry is the vertical line passing through the vertex. that is, the vertical line passing through (-2,-9)
so the equation for the axis of symmetry is: x = -2
last question on page 2
f(x) = x^4 - 3x^3 + 4x^2 - 6x + 4
we get f(0) when we replace x with 0
=> f(0) = 0^4 - 3(0)^3 + 4(0)^2 - 6(0) + 4
=> f(0) = 4
=> f(-4) = (-4)^4 - 3(-4)^3 + 4(-4)^2 - 6(-4) + 4
=> f(-4) = 256 + 192 + 64 + 24 + 4 = 540
1 is a zero of f if f(1) = 0
f(1) = 1^4 - 3(1)^3 + 4(1)^2 - 6(1) + 4
=> f(1) = 1 - 3 + 4 - 6 + 4 = 0
so yes, 1 is a zero of f(x)
since 1 is a zero of f(x), x - 1 is a factor of f(x)
so we divide x - 1 into f(x) by long division or synthetic division, either is too hard to type here, so i hope you know how to do that. we obtain:
f(x) = x^4 - 3x^3 + 4x^2 - 6x + 4 = (x - 1)(x^3 - 2x^2 + 2x - 4)
...............................................= (x - 1)[x^2(x - 2) + 2(x - 2)]
...............................................= (x - 1)[(x - 2)(x^2 + 2)]
...............................................= (x - 1)(x - 2)(x^2 + 2)
so the zeroes are given by:
x - 1 = 0
=> x = 1
or
x - 2 = 0
=> x = 2
or
x^2 + 2 = 0
=> x^2 = -2
...no solution, we get an imaginary number here
so all the zeroes are: x = 1, and x = 2