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3 years ago

11

40 Points correct answer

1 answer:

TiliK225 [7]3 years ago

6 0

<span>-2x^3(8x-7)+4x^3(3x-3)
= -16x^4 + 14x^3 + 12x^4 - 12x^3
= - 4x^4 + 2x^3

hope it helps</span>

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The ruler
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Rewrite as y−k=a(x−h)2 or x−h=a(y−k)2. Find the vertex, focus, and directrix.<br> y−3=(2−x)^2

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Given:

The given equation is

$y-3=(2-x)^2$

To find:

The vertex, focus, and directrix.

Solution:

The equation of a parabola is

$y-k=a(x-h)^2$ ...(i)

where, (h,k) is vertex, $\left(h,k+\dfrac{1}{4a}\right)$ and directrix is $y=k-\dfrac{1}{4a}$

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It can be written as

$y-3=(-(x-2))^2$

$y-3=(x-2)^2$ ...(ii)

On comparing (i) and (ii), we get

$h=2,k=3,a=1$

Vertex of the parabola is (2,3).

$Focus=\left(2,3+\dfrac{1}{4(1)}\right)$

$Focus=\left(2,3+\dfrac{1}{4}\right)$

$Focus=\left(2,\dfrac{13}{4}\right)$

Therefore, the focus of the parabola is $\left(2,\dfrac{13}{4}\right)$ .

Directrix of the parabola is

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3 years ago

Consider the sequence: 3, 8, 13, 18, 23, ...The recursive formula for this sequence is:an = an-1 + 5In a COMPLETE sentence, expl

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Answer:

see the explanation

Step-by-step explanation:

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In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant called the common difference

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Let

$a_1=3\\a_2=8\\a_3=13\\a_4=18\\a_5=23\\$

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$a_3-a_2=13-8=5$ ------> $a_3=a_2+5$

$a_4-a_3=18-13=5$ ------> $a_4=a_3+5$

$a_5-a_4=23-18=5$ ------> $a_5=a_4+5$

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For n > 1

where

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a(n-1) is the known term position (n-1)

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$a_8=a_(_8_-_1_)+5$

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or

we know that

The general formula for arithmetic sequence is equal to

$a_n=a_1+d(n-1)$

where

an is the term that you want to find (position n)

a_1 is the first term

d is the common difference

n is the number of terms

we have

$a_1=3$

$d=5$

substitute

$a_n=3+5(n-1)$

$a_n=3+5n-5$

$a_n=5n-2$

so

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