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valkas [14]
3 years ago
7

Match the function with its value f(x)= x2-3x+3 g(x)= 2x2-1 f(z^2)= .....

Mathematics
1 answer:
Elis [28]3 years ago
3 0
There's nothing here to 'match' it with.

f(x) = x^2 - 3x + 3

f(z^2) = (z^2)^2 - 3z^2 + 3

f(z^2) = z^4 - 3z^2 + 3

<em><u>g(x)</u></em> is only there to confuse you.
Apparently it was successful.
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given that BD is the median of ABD and that ABD is isosceles congruence postulate SSS can be used to prove which of the followin
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It is given in the question that

BD is the median. So it divides the opposite sides in two equal parts .

Therefore in triangles BAD and BCD,

AB and AC are congruent because of isosceles triangle.

AD and CD are congruent because of the median BD.

And BD and BD are congruent .

So the two triangles are congruent by SSS and the correct option is the first option .

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Step-by-step explanation:

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Is domain of f(x,y) = 1 + (4 -y^2)^1/2 open, closed or neither<br><br> is it bounded or unbounded?
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Closed and Bounded.

Step-by-step explanation:

Hi there!

1) Let's start by finding the values in which the function is defined. Remember that this can be rewritten:

f(x,y) = 1 + (4 -y^2)^{\frac{1}{2}} \:id\:est\:f(x,y)=1+\sqrt{(4 -y^2)}

Since every quadratic root are defined for values \geq 0  then, this help us to understand that we need calculate what interval this Domain is:

4-y^{2} \geq 0\\4-y^{2}-4\geq -4\\-y^{2}\geq-4 \\y^{2}\leq4\\-2\leq y\leq 2

D=[-2,2]

2) Graphically speaking, the domain is closed. For the values -2 and 2 are included, and bounded.

Bounded functions have all of their points contained by some circle origin centered. Check it out below.

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