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2 years ago

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Consider the function represented by the equation y-6x-9=0. Which answer shows the equation written in function notation with x

as the independent variable? A.f(x)= 6x+9 B. f(x)=1/6x+3/2 C. f(y)=1/6y+3/2

2 answers:

spin [16.1K]2 years ago

6 0

we have

$y-6x-9=0$

Step $1$

<u>Clear the variable y</u>

$y-6x-9=0$

Adds $(6x+9)$ both sides

$y-6x-9+(6x+9)=0+(6x+9)$

$y=(6x+9)$

Step $2$

<u>Convert in function notation</u>

Let

$f(x)=y$

$f(x)=(6x+9)$

therefore

<u>the answer is</u>

$f(x)=(6x+9)$

Nina [5.8K]2 years ago

4 0

Solution:

There are two kind of variables when equation of a function is written

1. Independent variable

2. Dependent Variable

A variable is said to be independent , if it can select constant values by itself.

And, a variable is said to be independent, if it is dependent on other variable of the equation.

The given equation is , y-6 x -9=0

As, we have to write a function in which x is an independent variable and other variable y, will work as a dependent variable.

So, writing the above linear function as,

y= 6 x + 9

f(x)= 6 x + 9 →→Option (A)

Correct representation of function in which x is an independent variable and y is a dependent variable,because if we substitute different real values of x, we get different real values of y.

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what is a correct first step in solving the inequality –4(3 – 5x)≥ –6x 9? –12 – 20x ≤ –6x 9 –12 – 20x ≥ –6x 9 –12 20x ≤ –6x 9 –1

DIA [1.3K]

<span>The answer is -12 + 20x ≥ –6x 9. The inequality is –4(3 – 5x)≥ –6x 9. The first step is to multiply factors on the left side. The intermediary steps are: (-4)*(3) - (-4)*(5x) ≥ –6x 9. -12 - (-20x) ≥ –6x 9. So, the first step will be: -12 + 20x ≥ –6x 9.</span>

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2 years ago

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Express these numbers in standard form (a)5730(b)0.23(c)0.000231(d)7150000(e)500.8

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Answer:

5730 = 5.73×10³

0.23 = 2.3×10‐¹

0.000231 = 2.31×10-⁴

7150000 = 7.15×10⁶

500.8 = 5.008×10²

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2 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.

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