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Semenov [28]
2 years ago
14

HELP ME PLSSSSSSSSSSS

Mathematics
1 answer:
Trava [24]2 years ago
8 0

Answer:

8

Step-by-step explanation:

I believe that the answer is 8

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Question 1 of 10
Nutka1998 [239]

Answer:

F) 15/8

Step-by-step explanation:

<em>A. 17/15</em>

<em>B. 8/17</em>

<em>C. 15/17</em>

<em>D. 8/15</em>

<em>E. 17/8</em>

<em>F. 15/8</em>

8 0
2 years ago
Can you help and tell me how you got the answer
geniusboy [140]
Y= -2X
Y= -2•3
Y= -6
X=3

Y= -2X
Y= -2•(-3)
Y=6
X= -3

Answer is y= -2X
8 0
3 years ago
(2x+2y)+(3y-5x) EIHOIVHNVIHVIOE
Nina [5.8K]

\huge\textsf{Hey there!}

\large\text{(2x + 2y) + (3y - 5x)}

\rightarrow\large\text{2x + 2y + 3y - 5x }

\large\textsf{COMBINE the LIKE TERMS}

\large\text{(2x - 5x) + (2y + 3y)}

\large\text{2x - 5x = \boxed{\bf -3x}}

\large\text{-3x + (2y + 3y)}

\large\text{2y + 3y = \boxed{\bf 5y}}

\large\text{\bf -3x + 5y}

\boxed{\boxed{\large\textsf{Answer: \huge \bf -3x + 5y}}}\huge\checkmark

\large\textsf{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

4 0
3 years ago
Read 2 more answers
State the domain and range of both the function and the inverse function. f(x) = 8 x + 8 f -1(x) = Domain of f(x): Range of f(x)
Katena32 [7]

Answer:

f^{-1}(x) = \frac{x+8}{8}

Domain of f(x): All real values.

Range of f(x): All real values.

Domain of f -1(x): All real values.

Range of f -1(x): All real values

Step-by-step explanation:

We are given the following function:

f(x) = 8x + 8

Since it is a line, we have that both the domain is the range is the real set.

Inverse function:

The domain of the inverse function is the range of the original function, and the range of the inverse function is the domain of the original function.

Finding the inverse function:

Exchange x and y in the original function, and isolate y. So

Original

y = 8x + 8

Exchanging:

x = 8y - 8

Isolating y:

8y = x + 8

y = \frac{x+8}{8}

f^{-1}(x) = \frac{x+8}{8}

5 0
3 years ago
I need assistance please help
Karo-lina-s [1.5K]
The answer is the third one need any more help let me know
6 0
3 years ago
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