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max2010maxim [7]
2 years ago
5

A. The first five terms of n^2 + 5 are

Mathematics
1 answer:
san4es73 [151]2 years ago
3 0

Given:

The nth term of a sequence is:

n^2+5

To find:

The first five terms of the given sequence.

Solution:

The given sequence is:

a_n=n^2+5

For n=1,

a_1=1^2+5

a_1=1+5

a_1=6

For n=2,

a_2=2^2+5

a_2=4+5

a_2=9

For n=3,

a_3=3^2+5

a_3=9+5

a_3=14

For n=4,

a_4=4^2+5

a_4=16+5

a_4=21

For n=5,

a_5=5^2+5

a_5=25+5

a_5=30

Therefore, the first five terms of the given sequence are 6, 9, 14, 21, 30.

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Given y = a(x - h)^2 + k, p = <br> A. a <br> B. 4a <br> C. 1/4a
Flauer [41]

9514 1404 393

Answer:

  C.  1/(4a)

Step-by-step explanation:

We assume you're comparing the vertex form ...

  y = a(x -h)^2 +k

to the form used to write the equation in terms of the focal distance p.

  y = 1/(4p)(x -h)^2 +k

That comparison tells you ...

  a = 1/(4p)

  p = 1/(4a) . . . . . . multiply by p/a; matches choice C

__

<em>Additional comment</em>

When using plain text to write a rational expression, parentheses are needed around any denominator that has is more than a single constant or variable. The order of operations requires 1/4a to be interpreted as (1/4)a. The value of p is 1/(4a).

When rational expressions are typeset, the fraction bar serves as a grouping symbol identifying the entire denominator:

  p=\dfrac{1}{4a}

8 0
3 years ago
Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
Lorico [155]

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

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3 years ago
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