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nadya68 [22]
2 years ago
13

What are the benefits of creating and using functions in your code?.

Mathematics
1 answer:
ch4aika [34]2 years ago
4 0
It is critical to remember that creating functions while coding is a vital component of coding. There are various benefits to doing so.
A few of the advantages include the following:

The function can be called from anywhere in the code and made to perform exactly what it was written for the first time if it is called from a different line of code.

2. Writing functions reduces redundancy by allowing for fewer lines of code to be written in total. It eliminates the need to rewrite numerous lines of code over and over again. After the initial line of code has been set, we can call the method from another location.

3. It increases the efficiency of coding.

4. It makes it easier for others to comprehend the fundamentals involved in a coding structure, which is an advantage.
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ruslelena [56]
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6 0
2 years ago
5x-6=3x<br> what does x equal
makvit [3.9K]
5x-6= 3x
⇒ 5x-3x= 6
⇒ 2x= 6
⇒ x= 6/2
⇒ x= 3

Final answer: x=3~
3 0
3 years ago
Read 2 more answers
The area of a rectangular garden can be expressed as A = 2x^2 -x -6. The width of the garden is 2x+3. Find an expression for the
zalisa [80]

Answer:

Step-by-step explanation:

2x² -x - 6 = 2x² - 4x + 3x - 3*2

              = 2x*(x - 2) + 3*(x - 2)

              = (x-2)(2x + 3)

A = 2x² -x - 6

A = (x - 2)(2x + 3)

length*width = (x - 2)(2x + 3)

length * (2x +3) = (x - 2)(2x + 3)

length=\frac{(x-2)(2x+3)}{(2x+3)}

length = x - 2

5 0
3 years ago
What is the surface area of this design? 5 in 5 in 6.4 in 5 in 9 in
xeze [42]

ANSWER

197 {in}^{2}

EXPLANATION

The area of the two tra-pezoidal faces

= 2 \times  \frac{1}{2}  \times (9 + 5) \times 5 = 70 {in}^{2}

The area of the 5 by 6.4 rectangular face

= 6.4 \times 5 = 32  {in}^{2}

The area of the two square faces

= 2(5 \times 5) = 50 {in}^{2}

The area of the 9 by 5 rectangular face is

= 9 \times 5 = 45 {in}^{2}

The surface area of the design is:

70 + 32 + 50  + 45=197 {in}^{2}

6 0
3 years ago
What is the GCF of 44j5k4 and 121j2k6?<br><br> 4j3k2<br> 4j2k4<br> 11j3k2<br> 11j2k4
egoroff_w [7]

Answer:

D) 11j2k4

Step-by-step explanation:

Simply find the prime factors of each term!

44 j5 k4

121 j2 k6

11 goes into 44 and 121. So, we can use 11!

2 works for j5 and j2, because of 10, so you would input j2.

4 works for both k4 and 46 because of 24, so k4.

11 j2 k4

Hope this helps! :)

5 0
3 years ago
Read 2 more answers
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