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Oliga [24]
3 years ago
6

Can anyone please explain this?

Mathematics
1 answer:
Leni [432]3 years ago
4 0

Answer:

Solution given:

Let there be a point P(x, y) equidistant from

A(-3, 2) and B(0,4),

so PA = PB,

\sqrt{(x+3)²+(y-2)²}=\sqrt{(x-0)²+(y-4)²}

squaring both side

(\sqrt{(x+3)²+(y-2)²})^{2}=(\sqrt{(x-0)²+(y-4)²})²

x²+6x+9+y²-4y+4=x²+y²-8y+16

x²+6x+y²-4y-x²-y²+8y=16-4-9

6x-4y+8y=3

<u>6x-4y=3 </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>r</u><u>e</u><u>q</u><u>u</u><u>i</u><u>r</u><u>e</u><u>d</u><u> </u><u>l</u><u>o</u><u>c</u><u>u</u><u>s</u><u> </u>

<u>A</u><u>c</u><u>t</u><u>u</u><u>a</u><u>l</u><u>l</u><u>y</u><u>:</u>

<u>A</u><u> </u><u>locus</u><u> </u><u>is</u><u> </u><u> </u><u>a curve or other figure formed by all the points satisfying a particular equation of the relation between coordinates, or by a point, line, or surface moving according to mathematically defined conditions.</u>

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<h3>Answer is   -9</h3>

=================================

Work Shown:

(g°h)(x) is the same as g(h(x))

So, (g°h)(0) = g(h(0))

Effectively h(x) is the input to g(x). Let's first find h(0)

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-------

alternatively, you can plug h(x) algebraically into the g(x) function

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g( h(x) ) = -3*( h(x) ) ... replace all x terms with h(x)

g( h(x) ) = -3*(x^2 + 3) ... replace h(x) on right side with x^2+3

g( h(x) ) = -3x^2 - 9

Next we can plug in x = 0

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g( h(0) ) = -9

we get the same result.

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