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DENIUS [597]
3 years ago
13

In the diagram, BC¯¯¯¯¯∥DE¯¯¯¯¯ . What is AE ?

Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0
So there are two triangles here: Smaller one (ADE) and bigger one (ABC) and they both are similar.

So you can use proportions here.

AB / AC = AD / AE

AB = AD + DB = 6 + 1 = 7
AC = AE + 3

AD = 6

So plug in these values:

AB / AC = AD / AE becomes
7 / (AE + 3) = 6 / AE

Now do the cross multiply:

7 AE = 6 (AE + 3)

Now solve for AE:

7AE = 6AE + 18

AE = 18
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2 years ago
Find a polynomial of degree 3 with real coefficients and zeros of -3,-1, and 4, for which f(-2)=-24
Zanzabum

We want to find a polynomial

<em>f(x)</em> = <em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em>

such that the roots of <em>f</em> are <em>x</em> = -3, <em>x</em> = -1, and <em>x</em> = 4, and <em>f(x)</em> takes on a value of -24 when <em>x</em> = -2.

The factor theorem for polynomials tells us that we can factorize <em>f(x)</em> as

<em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em> = <em>a</em> (<em>x</em> + 3) (<em>x</em> + 1) (<em>x</em> - 4)

Expand the right side:

(<em>x</em> + 3) (<em>x</em> + 1) (<em>x</em> - 4) = <em>x</em>³ - 13<em>x</em> - 12

So we have

<em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em> = <em>a x</em>³ - 13<em>a</em> <em>x</em> - 12<em>a</em>

<em />

In order for both sides to be equal, the coefficients of both polynomials on terms of the same degree must be equal. This means

<em>a</em> = <em>a</em> (of course)

<em>b</em> = 0 (there is no <em>x</em>² term on the right)

<em>c</em> = -13<em>a</em>

<em>d</em> = -12<em>a</em>

<em />

We also have that <em>f</em> (-2) = -24, which means

<em>f</em> (-2) = <em>a</em> (-2 + 3) (-2 + 1) (-2 - 4)

-24 = 6<em>a</em>

<em>a</em> = -4

which in turn tells us that <em>c</em> = 52 and <em>d</em> = 48.

So we found

<em>f(x)</em> = -4<em>x</em>³ + 52<em>x</em> + 48

4 0
2 years ago
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