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3 years ago

Decompose 8/12 in two ways with: x/2 + x/2 = 8/12

1 answer:

8 0

$\frac{x}{2} + \frac{x}{2} = \frac{8}{12} \\\frac{2x}{2} = \frac{2}{3} \\x = \frac{2}{3}$
or
$\frac{x}{2} + \frac{x}{2} = \frac{8}{12} \\\frac{x}{2} + \frac{x}{2} = \frac{2}{3} \\\frac{\frac{2}{3}}{2} + \frac{\frac{2}{3}}{2} = \frac{2}{3} \\\frac{1}{3} + \frac{1}{3} = \frac{2}{3} \\\frac{2}{3} = \frac{2}{3}$

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3 years ago

Captain John Smith earns $3,300 a month in the Air Force. If the Air Force puts 6% into his pension, how much goes into his pens

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3 years ago

Maple street is 3 times as long as pine street . how long is pine street?

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3 years ago

Read 2 more answers

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

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

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

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 $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain 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.

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: $y = \sqrt [3] {2}$ in the same way is defined.

Answer:

$y = \sqrt {x-2}$

Option b

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4 years ago

HELP PLSS I'M LOST PLSSSSSSSSSSSSSSSSS

Svetach [21]

Answer:

First Row:

75/100

0.75

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Second Row:

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10%

Step-by-step explanation:

Count the squares that are shaded in out of the 100.

Just divide the fraction together for the decimal.

Move the decimal two places to the right to get the percentage

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3 years ago