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Andreyy89
3 years ago
15

Each statement describes a transformation of the graph of y = log2x.

Mathematics
2 answers:
egoroff_w [7]3 years ago
3 0
I take it this is in base 2? It really doesn't matter for this question, but I will interpret it that way. The easiest way to do this is with a graphing calculator to see what happened. You can do it on the internet by going to wolframalpha and putting the equations in like this.

y = log_2(x); y = log_2(x + 3) - 9

You will see the picture clearly when you do this. The answer is that it moves 3 units towards the left and 9 units down. 

Answer: Third one down.

Comment: Anything you put inside the brackets with an x moves the graph left or right depending on the sign. (x + 3) moves 3 units to the left. (x - 3) moves the graph 3 units to the left.

Anything outside the brackets moves the graph up or down.  minus goes down. Plus goes up. The number must be on the right. - 9 goes nine units down.
garri49 [273]3 years ago
3 0

it's C: It is the graph of y = log2x translated 9 units down and 3 units to the left.


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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,
Eva8 [605]

Answer:

Elements are of the form

 (i) [a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}

(ii) [a,b)=\{[x,y) :a\leq x

(iii)(a,a]=\{(x,y] :a

(iv)(a,a)=\{(x,y): a

(v) (a,b) where a>b=\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}

(vi) [a,b] where a>b=\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}

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.

\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R 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.

[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if a\in \mathbb R 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.

( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R 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.

( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Because there is no increment so if a\in \mathbb R 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,

(a,b)=\{(5,0),(5,1),(5,2).....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

(vi) [a,b] where a\leq b.

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,

[a,b]=\{[5,0],[5,1],[5,2].....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

8 0
3 years ago
3q(p-q)+2r(p-q)-S(q-p)<br> Can u solve it step by step
Artyom0805 [142]

Answer:

-3q² + 3qp + 2rp - 2rq + Sq - Sp

Step-by-step explanation:

first part

3q(p-q) = 3qp - 3q²

second part

2r(p-q) = 2rp - 2rq

third part

S(q-p) = Sq - Sp

then we put it all together

3qp - 3q² + 2rp - 2rq + Sq - Sp

in the right place possibly

-3q² + 3qp + 2rp - 2rq + Sq - Sp

8 0
2 years ago
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