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Natali5045456 [20]
3 years ago
7

Find (ƒ + g)(x) for the functions provided: ƒ(x) = 2x − 5, g(x) = 4x + 3

Mathematics
2 answers:
svetoff [14.1K]3 years ago
8 0

Answer:  The correct option is

(A) (ƒ + g)(x) = 6x − 2.

Step-by-step explanation:  We are given to find the value of (f + g)(x) for the following functions :

f(x)=2x-5,\\\\g(x)=4x+3.

We know that

for any two functions p(x) and q(x), we have

(f+g)(x)=p(x)+q(x).

So, we get

(f+g)(x)\\\\=f(x)+g(x)\\\\=2x-5+4x+3\\\\=6x-2.

Thus, (f+g)(x)=6x-2.

Option (A) is CORRECT.

Ann [662]3 years ago
4 0
This is 4x + 3 + 2x - 5

= 6x - 2 answer
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Semmy [17]

Answer:

x = 14

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Step-by-step explanation:

`From the given picture,

Angle SOP and angle SOR are the linear pairs.

Therefore, sum of these angles will be equal to to 180°.

m∠SOP + m∠SOR = 180° -----(1)

Since, m∠SOP = (7x - 2)° and m∠SOR = (5x + 14)°

[There is a misprint in this question. There should be x in place of y in the measures of the angles]

By substituting these values in the equation (1),

(7x - 2) + (5x + 14) = 180

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5 0
3 years ago
Helppppppppppp:)))))))))
Whitepunk [10]

Hi there!

We are given the set of ordered pairs below:

\large \boxed{(3, - 1),(2, - 2),(0,2),(2,1)}

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!

4 0
3 years ago
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