Answer:
All Real Numbers except for 4
Step-by-step explanation:
Objective: Specify domain and range of a function.
The equation is a linear equation. A linear equation range is all real numbers. However x cannot equal 1 so let plug in x to find where the y value CANT be.
So the range is
All Real Numbers except for 4.
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.
Answer:
x = 2√2
y = 2√6
Step-by-step explanation:
Consider the ratio of the two legs of the two smaller interior right triangles. (refer to attached diagrams for the triangles - I have outlined one in blue and the other in red)
These will be equal since the triangles are similar
shorter leg : longer leg
(blue triangle = red triangle)
⇒ x : 4 = 2 : x
⇒ x/4 = 2/x
⇒ x² = 8
⇒ x = √8
⇒ x = 2√2
Now we have x, we have the two legs of the right triangle with hypotenuse labelled y.
Using Pythagoras' Theorem a² + b² = c² (where a and b are the legs and c is the hypotenuse)
⇒ 4² + (2√2)² = y²
⇒ y² = 24
⇒ y = √24
⇒ y = 2√6
