Given: ⅓ (18x + 27) = 81 Prove: x = 12
1 answer:
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Answer:
see attached
Step-by-step explanation:
I find it convenient to let a graphing calculator draw the graph (attached).
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If you're drawing the graph by hand, there are a couple of strategies that can be useful.
The first equation is almost in slope-intercept form. Dividing it by 2 will put it in that form:
y = 2x -4
This tells you that the y-intercept, (0, -4) is a point on the graph, as is the point that is up 2 and right 1 from there: (1, -2). A line through those points completes the graph.
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The second equation is in standard form, so the x- and y-intercepts are easily found. One way to do that is to divide by the constant on the right to get ...
x/2 +y/3 = 1
The denominators of the x-term and the y-term are the x-intercept and the y-intercept, respectively. If that is too mind-bending, you can simply set x=0 to find the y-intercept:
0 +2y = 6
y = 6/2 = 3
and set y=0 to find the x-intercept
3x +0 = 6
x = 6/3 = 2
Plot the intercepts and draw the line through them for the graph of this equation.
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Here, we have suggested graphing strategies that don't involve a lot of manipulation of the equations. The idea is to get there as quickly as possible with a minimum of mistakes.
Answer:
Elements are of the form
(i)
(ii)
(iii)
(iv)
(v) (a,b) where a>b=
(vi) [a,b] where a>b=
Step-by-step explanation:
Given intervals are,
(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.
To show all its elements,
(i) [a,a]
Imply the set including aa from left as well as right side.
Its elements are of the form.
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].
(ii) [a,a)
This means given interval containing a by left and exclude a by right.
Its elements are of the form.
Since there is a singleton element a of real numbers withis the set, this set is empty.
Because there is no increment so if then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).
(iii) (a,a]
It means the interval not taking a by left and include a by right.
Its elements are of the form.
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].
(iv) (a,a)
Means given set excluding a by left as well as right.
Since there is a singleton element a of real numbers, this set is empty.
Its elements are of the form.
Because there is no increment so if then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).
(v) (a,b) where a>b.
Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.
That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,
e.t.c
So this set is connected and we know singletons are connected in . Hence given set is empty.
(vi) [a,b] where .
Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.
That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,
e.t.c
So this set is connected and we know singletons are connected in . Hence given set is empty.