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1 year ago

What is the domain of f(x) = (x + 2)2 – 1?

Mathematics

1 answer:

larisa [96]1 year ago

4 0

Simplify

$\\ \tt\Rrightarrow f(x)=(x+2)^2-1$

$\\ \tt\Rrightarrow f(x)=x^2+4x+2-1$

$\\ \tt\Rrightarrow f(x)=x^2+4x+1$

The Domain is set of all real numbers

$\\ \tt\Rrightarrow Domain\in R$

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Help, plz............

Brilliant_brown [7]

Answer:

I think Question (a) is 11748 m and Question (b) is 32364

Step-by-step explanation:

Because I Just divided the number of 7031 m square by 79 in Question (a) and I Just multiplied the number of 348 m by 93 but I am not sure It's correct or not

I hope It's helpful

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2 years ago

If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No

Effectus [21]

If A and B are independent, then $P(A\cap B)=P(A)P(B)$ .

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$

$P(A\cup B)=0.5+0.2-0.5\cdot0.2$

$\boxed{P(A\cup B)=0.6}$

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

$P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)$

$P(A^c\cap B^c)=1-0.6$

$\boxed{P(A^c\cap B^c)=0.4}$

c. DeMorgan's law can be used here too:

$P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)$

$P(A^c\cup B^c)=1-0.5\cdot0.2$

$\boxed{P(A^c\cup B^c)=0.9}$

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2 years ago

How many hours is 10 percent of 24 hours

Ne4ueva [31]

It is 2.4 hours I hope this helped! :)

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romanna [79]

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3 years ago

Triangle ABC is circumscribed about Circle O. Use the diagram below to describe Point O.

Gwar [14]

Since this is an obtuse triangle, Point O is not equidistant from A, B, and C. Point O is not on the perpendicular bisectors, so the third statement is true. Point O is equidistant from AB, BC, and CA because these lines are pressed against the circpe in a mannered way.

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2 years ago

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