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Sedbober [7]
3 years ago
6

In an arithmetic sequence if a1 =7 and d = 3, find a64

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
6 0

Answer:

a_{64}=196

Step-by-step explanation:

<u>Arithmetic sequence</u>

General form of an arithmetic sequence: a_n=a+(n-1)d

where:

  • a_n is the nth term
  • a is the first term
  • d is the common difference between terms

Given:

  • a_1=7
  • d=3

Substituting these into the equation to find the nth term:

\implies a_n=7+(n-1)3

\implies a_n=3n+4

To find the 64th term, substitute n = 64 into the formula:

\implies a_{64}=3(64)+4=196

masya89 [10]3 years ago
5 0

Answer:

<u>196</u>

Step-by-step explanation:

<u>Formula for the nth term</u>

  • aₙ = a₁ + (n - 1) d

<u>Taking n = 64</u>

  • a₆₄ = 7 + 63(3)
  • a₆₄ = 7 + 189
  • a₆₄ = <u>196</u>
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3 years ago
E QUESTION ONE<br> Solve the proportion—Show all work.<br> 5<br> x 1 W<br> =<br> 65<br> X Х<br> XX
Lena [83]

Answer:

x=39

Step-by-step explanation:

3/x  = 5/65

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7 0
3 years ago
Read 2 more answers
Can someone help answer the first two?
Murljashka [212]

Answer:

below given!

Step-by-step explanation:

1) mean = (7.4 + 5.8 + 7.3 + 7 + 8.9 + 9.4 + 8.3 + 9.3 + 6.9 + 7.5 + 9 + 5.8 + 5.5 + 8.6 + 9.3 +3.8) / 16

mean = 7.47

5 0
3 years ago
Solve for x. Simplify completely.<br> x^2+4x+4=8
velikii [3]
X^2+4x+4=8
-8                -8
subtract 8 from both sides 

x^2+4x-4=0

now factor using quadratic formula
ax^2+bx+c=0

x= -b +-[ (sqrt (b^2-4ac))/2a ]

x=-2(1+sqrt(2))


x= 2(sqrt(2) -1)

hope that helps, hopefully you know how to plug and chug into quadratic formula. if not you can ask for help


5 0
4 years ago
What is the solution of -8/2y-8=5/y+4-7y+8/y^2-16
Mnenie [13.5K]
<h2>Answer:</h2>

The solution of the given equation is:

                         y=6

<h2>Step-by-step explanation:</h2>

The expression is given by:

\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}

Now on solving for the given equation

\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}

since,

a^2-b^2=(a-b)(a+b)\\\\so,\\\\y^2-16=y^2-4^2\\\\i.e.\\\\y^2-16=(y-4)(y+4)

Hence, we get:

\dfrac{-4}{y-4}=\dfrac{5\times (y-4)-(7y+8)}{(y+4)(y-4)}\\\\i.e.\\\\\dfrac{-4}{y-4}=\dfrac{5y-20-7y-8}{(y-4)(y+4)}

-4\times (y-4)(y+4)=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4\times (y+4))=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4y+16)=(-2y-28)(y-4)\\\\i.e.

(y-4)(-4y+16)-(-2y-28)(y-4)=0\\\\i.e.\\\\(y-4)(-4y-16+2y+28)=0\\\\i.e.\\\\(y-4)(-2y+12)=0\\\\i.e.

y-4=0\ or\ -2y+12=0\\\\i.e.\\\\y=4\ or\ 2y=12\\\\i.e.\\\\y=4\ or\ y=6

but y≠ 4

since, the denominator of the term in the left side of the given equality and the second term in the right side of the given equality will be zero and hence, the expression will be not defined.

Hence, the value of y is: 6

7 0
4 years ago
Read 2 more answers
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