Answer : 96
x – y = 16 --------> equation 1
x is the higher grade and y is the lower grade
We solve the first equation for y
x - y = 16
-y = 16 -x ( divide each term by -1)
y = -16 + x
Now substitute y in second equation
Take common denominator to combine fractions
Add 8 on both sides
Multiply both sides by
x = 96
We know x is the higher grade
96 is the higher grade of Jose’s two tests.
By using the picture that was provided below, you can see that
m<MXN is greater than m<YZX.
Angles m<LZX is equal to m<ZYX + m<YZX and
m<MYL is equal to 180 degrees minus M<ZYX.
I hope this helps! Have a good day
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.