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3 years ago

Determine the equation of the midline of the following graph. ITS NOT HARD I JUSST DONT KNOW IT!!!!HELP PLEASE.

Mathematics

1 answer:

photoshop1234 [79]3 years ago

6 0

Answer:

y=1

Step-by-step explanation:

it is in the middle of the maximum and minimum.

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A real estate firm charges 2% commission fee on sales. If Maha sold her house and was charged $13,400 for commission, what was t

Vinil7 [7]

Answer:

$670,000

Step-by-step explanation:

13,400 multiplied by 50 is 670,000

5 0

4 years ago

There were several sunny days this year before my birthday. Since my birthday, there have been 171717 more sunny days. My birthd

Lesechka [4]

Answer:

66 sunny days before birthday

Step-by-step explanation:

Total sunny days after birthday = 17

Total sunny days in year = 83

Birthday was not sunny

Total sunny days before birthday = 83 -17

= 66 sunny days

Read more on Brainly.in - https://brainly.in/question/6937259#readmore

4 0

4 years ago

The function g(x)=x^2+3. The f(x)=g(x+2)

vekshin1

Answer:

$f(x)=x^2+6x+12$

Step-by-step explanation:

Given:

The function 'g(x)' is given as:

$g(x)=x^2+3$

Now, the function 'f(x)' is given as:

$f(x)=g(x+2)$

So, the function 'f(x)' is a transformation of 'g(x)'.

In order to find f(x), we replace 'x' by $(x + 2)$ in the 'g(x)' function equation. This gives,

$g(x+2)=(x+2)^2+3$

Using the identity $(a+b)^2=a^2+b^2+2ab$ , we get:

$g(x+2)=x^2+3^2+6x+3\\\\g(x+2)=x^2+9+6x+3\\\\g(x+2)=x^2+6x+12$

Hence the function f(x) is given as:

$f(x)=x^2+6x+12$

4 0

4 years ago

A 2-column table with 7 rows. Column 1 is labeled Power of 3 with entries 3 cubed, 3 squared, 3 Superscript 1, 3 Superscript 0,

Masja [62]

Answer:

a=1, b=1/3, c=1/27

Step-by-step explanation:

The table is given below:

$\left|\begin{array}{c|c}$Powers of 3&Value\\----&---\\3^3&27\\3^2&9\\3^1&3\\3^0&a\\3^{-1}&b\\3^{-2}&\frac{1}{9} \\3^{-3}&c\end{array}\right|$

(a)As the exponents decrease, the next term is gotten by a division of the previous term by 3.

$f_n=f_{n-1}\div 3$

(b)

$f_n=f_{n-1}\div 3\\I.a=f_n, f_{n-1}=3,\\a=f_n=3\div 3=1\\II. b=f_n, f_{n-1}=1,\\b=f_n=1\div 3=\frac{1}{3}\\III. c=f_n, f_{n-1}=\frac{1}{9},\\c=f_n=\frac{1}{9}\div 3=\frac{1}{27}$

a=1, b=1/3, c=1/27

7 0

3 years ago

Read 2 more answers

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

OleMash [197]

Answer:

that makes my head hurt n o

Step-by-step explanation:

AAAAAAA

6 0

3 years ago