2(x + 3) = 18
2x + 6 = 18
- 6 =
2x = 12
Divide 2
x = 6
The common denominator is 12
Answer:
Simplifying
5x2 + -10x = 2
Reorder the terms:
-10x + 5x2 = 2
Solving
-10x + 5x2 = 2
Solving for variable 'x'.
Reorder the terms:
-2 + -10x + 5x2 = 2 + -2
Combine like terms: 2 + -2 = 0
-2 + -10x + 5x2 = 0
Begin completing the square. Divide all terms by
5 the coefficient of the squared term:
Divide each side by '5'.
-0.4 + -2x + x2 = 0
Move the constant term to the right:
Add '0.4' to each side of the equation.
-0.4 + -2x + 0.4 + x2 = 0 + 0.4
Reorder the terms:
-0.4 + 0.4 + -2x + x2 = 0 + 0.4
Combine like terms: -0.4 + 0.4 = 0.0
0.0 + -2x + x2 = 0 + 0.4
-2x + x2 = 0 + 0.4
Combine like terms: 0 + 0.4 = 0.4
-2x + x2 = 0.4
The x term is -2x. Take half its coefficient (-1).
Square it (1) and add it to both sides.
Add '1' to each side of the equation.
-2x + 1 + x2 = 0.4 + 1
Reorder the terms:
1 + -2x + x2 = 0.4 + 1
Combine like terms: 0.4 + 1 = 1.4
1 + -2x + x2 = 1.4
Factor a perfect square on the left side:
(x + -1)(x + -1) = 1.4
Calculate the square root of the right side: 1.183215957
Break this problem into two subproblems by setting
(x + -1) equal to 1.183215957 and -1.183215957.
Step-by-step explanation:
Answer:
Sure.
Step-by-step explanation:
easy answer
yes x no = sure
Sure x maybe = Ok
Hi there!
We are given the set of ordered pairs below:
1. What is the domain?
- Domain is a set of all x-values in one set of ordered pairs. So what are the x-values that I am talking about? In ordered pairs, we define x and y which both have relation to each others which we can write as (x,y). That's right, the domain is set of all x-values from ordered pairs.
Therefore, we gather only x-values from (x,y). Hence, the domain is {3,2,0,2}. Whoops! Something is not right. As we learn in Set Theory that we don't write the same or repetitive in a set. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>c</u><u>t</u><u>u</u><u>a</u><u>l</u><u> </u><u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>0</u><u>,</u><u>2</u><u>,</u><u>3</u><u>}</u>
2. What is the range?
- Because domain is set of all x-values. Then what do you think the range is? That's right! The range is <u>s</u><u>e</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>l</u><u>l</u><u> </u><u>y</u><u>-</u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>s</u><u>.</u> If you got this right before looking up the underlined words then a handclap for you! So how do we find range? Simple, we just do like finding the domain in the Q1, except we gather the y-values in (x,y) instead and make sure that we don't write same number!
Therefore, gather y-values from the ordered pairs. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>r</u><u>a</u><u>n</u><u>g</u><u>e</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>-</u><u>2</u><u>,</u><u>-</u><u>1</u><u>,</u><u>1</u><u>,</u><u>2</u><u>}</u>
3. Is the relation a function?
- All functions are relations but not all relations are functions. Function is a set of ordered pairs where <u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>n</u><u>o</u><u>t</u><u> </u><u>r</u><u>e</u><u>p</u><u>e</u><u>t</u><u>i</u><u>t</u><u>i</u><u>v</u><u>e</u><u> </u><u>o</u><u>r</u><u> </u><u>i</u><u>n</u><u> </u><u>a</u><u> </u><u>s</u><u>e</u><u>t</u><u>,</u><u> </u><u>t</u><u>h</u><u>e</u><u>r</u><u>e</u><u> </u><u>c</u><u>a</u><u>n</u><u>n</u><u>o</u><u>t</u><u> </u><u>b</u><u>e</u><u> </u><u>m</u><u>o</u><u>r</u><u>e</u><u> </u><u>t</u><u>h</u><u>a</u><u>n</u><u> </u><u>o</u><u>n</u><u>e</u><u> </u><u>s</u><u>a</u><u>m</u><u>e</u><u> </u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>.</u> Consider the following relation: (1,1),(1,2) - Oh, looks like in a set of ordered pairs, there are two same domains which make it only a relation, and not a function. On the other hand, (1,1),(2,2) - Looking good! No same or repetitive domain, making it indeed a function.
Consider the domain from Q1 and see if there are two same values of x in a set. Looks like the relation is not a function since there are same x-values which are 2 in a set, making it only a relation. Hence, the relation is not a function.
These are all 3 answers along with an explanation. Let me know if you have any doubts regarding Relations and Functions.
<em>F</em><em>r</em><em>o</em><em>m</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>Q</em><em>1</em><em>'</em><em>s</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>,</em><em> </em><em>t</em><em>h</em><em>e</em><em>r</em><em>e</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>t</em><em>w</em><em>o</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em>s</em><em>,</em><em> </em><em>p</em><em>l</em><em>e</em><em>a</em><em>s</em><em>e</em><em> </em><em>c</em><em>h</em><em>o</em><em>o</em><em>s</em><em>e</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em> </em><em>t</em><em>o</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>u</em><em>n</em><em>d</em><em>e</em><em>r</em><em>l</em><em>i</em><em>n</em><em>e</em><em>)</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>n</em><em>o</em><em>t</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>f</em><em>i</em><em>r</em><em>s</em><em>t</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>s</em><em>a</em><em>m</em><em>e</em><em> </em><em>2</em><em>'</em><em>s</em><em>)</em><em>.</em><em> </em>
Good luck on your assignment, have a nice day!