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2 years ago

Please help!!!!!!!!!!

Mathematics

2 answers:

Basile [38]2 years ago

6 0

Answer:

1-c

2-b

3-a

Step-by-step explanation:

1. 7y=14

y=2

-7y=-14

y=2

so the equation has one solution

2. 0=0

so the equation has infinite solution

3. 0=2 false

so the equation has no solution

Zanzabum2 years ago

3 0

1 solution

8x+7y=30

8x-7y=2

infinitely many solutions

-8x+7y=-2

8x-7y=2

no solution

8x-7y=2

-8x+7y=2

Explanation

The number of solutions of a pair of linear equations are determined by the following criteria

1 solution

- $a1/a2 \neq b1/b2$

in the first pair $a1/a2=8/8=1$

$b1/b2=7/-7=-1$

therefore $a1/a2 \neq b1/b2$

infinitely many solution

$a1/a2=b1/b2=c1/c2$

In the second pair of equations -8x+7y+2=0 and 8x-7y-2=0

$a1/a2=-8/8=-1\\b1/b2=7/-7=-1\\c1/c2=2/-2=-1$

since $a1/a2=b1/b2=c1/c2$ , the pair of equations have infinitely many solutions

no solution

$a1/a2=b1/b2\neq c1/c2$

in the third pair of equation

$a1/a2=8/-8=-1\\b1/b2=-7/7=-1\\c1/c2=-2/-2=1$

Here $a1/a2=b1/b2\neq c1/c2$

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3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

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For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .