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Novay_Z [31]
3 years ago
12

What is the domain of

Mathematics
2 answers:
Greeley [361]3 years ago
8 0

Answer:

All real numbers except 1

Anika [276]3 years ago
8 0

Answer:

All real numbers except 1

Step-by-step explanation:

The domain is the value that x can take from a given function. Therefore, we can not take the value of x as 1 otherwise the function will become undefined.

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Solve 7n 6 4n-2=2n 2(4n 6)
scoray [572]
More information is required to answer the question correctly.
6 0
3 years ago
PLZ HELP I GIVE 15 POINTS
Natali [406]

Answer:

<u>4 1/12. (Four, one by twelve).</u>

Step-by-step explanation:

7/12 × 7 =?

7/12×7 = <u>49/12.</u>

The answer is now in improper fraction. When we convert it to mixed fraction we get the answer as <u>4 1/12. (Four, one by twelve).</u>

8 0
3 years ago
A. 55<br> b. 70<br> c. 110<br> d. 125
morpeh [17]

Answer:

b

Step-by-step explanation:

7 0
3 years ago
The score on a trivia game is obtained by subtracting the number of incorrect answer from twice the number of correct answer. If
brilliants [131]

Answer:

The player answer correctly 30 questions

Step-by-step explanation:

First we have to develop the information that gives us as 2 equations

x = correct answer

y = incorrect answer

z = score = 40

h = total answer = 50

2x - y = z               x + y = h

2x - y = 40             x + y = 50

we clear the x of one equation and replace the value of x in the other equation

x = 50 - y

2x - y = 40

2(50 - y) - y = 40

100 - 2y - y = 40

-3y = 40 -100

y = -60/-3

y = 20

x = 50 - y

x = 50 - 20

x = 30

3 0
3 years ago
Suppose that a function​ f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence
Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

\lim_{x \to c} \frac{x^2-cx}{(x-c)}

We can simplify it

\lim_{x \to c} \frac{x(x-c)}{(x-c)}

=\lim_{x \to c} x

=c

so, we can see that limit exist and it's value defined

case-2:

\lim_{x \to c} \frac{1}{(x-c)}

Left limit is

\lim_{x \to c-} \frac{1}{(x-c)}

=-\infty

Right Limit is

\lim_{x \to c+} \frac{1}{(x-c)}

=+\infty

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0
3 years ago
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