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2 years ago

Graphing f(x)=(x+2)^2-3 in vertex form

Mathematics

1 answer:

USPshnik [31]2 years ago

5 0

Answer:

Step-by-step explanation:

This is a parabola shaped like a U.

The minimum value is at (-2, -3).

Find the zeroes:

(x + 2)^2 = 3

Find the zeroes:

x + 2 = +/- √3

x = +/-√3 - 2

x = -0.27, -3.73.

So the graph cuts the x axis at (-0.27, 0) and (-3.73, 0)

when x = -4 , f(x) = 1 and when x = 1, f(x) = 6.

So you can now draw the curve through these 5 points and it will be shaped like a U, symetrical about the line x = -2

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(x - y + z) : (y - z + 2w) : (2x + z - 10) = 2:3:5 Then ( 3x + 3z - 2w) : w = ?pgvzcmckyc?

sukhopar [10]

Answer:

Step-by-step explanation:

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There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

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where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

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By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

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$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

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

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2 years ago

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kakasveta [241]

Answer:

213 in base 4 = 124 in base 5

Step-by-step explanation:

First we convert base 4 to base 10

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