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Lilit [14]
3 years ago
11

In an arithmetic​ sequence, the nth term an is given by the formula An=a1+(n−1)d​, where a1

Mathematics
1 answer:
Gnesinka [82]3 years ago
5 0

Answer:

88

Step-by-step explanation:

We are given that in arithmetic sequence , the nth term a_n is given by the formula

A_n=a_1+(n-1)d

Where a_1=first term

d=Common difference

In an geometric sequence, the nth term is given by

a_n=a_1r^{n-1}

Where r= Common ratio

1,4,7,10,..

We have to find 30th term.

a_1=1,a_2=4,a_3=7,a_4=10

d=a_2-a_1=4-1=3

d=a_3-a_2=7-4=3

d=a_4-a_3=10-7=3

r_1=\frac{a_2}{a_1}=\frac{4}{1}=4

r_2=\frac{a_3}{a_2}=\frac{7}{4}

r_1\neq r_2

Therefore, given sequence is an arithmetic sequence because the difference between consecutive terms is constant.

Substitute n=30 , d=3 a=1 in the given formula of arithmetic sequence

Then, we get

a_{30}=1+(30-1)(3)=1+29(3)=1+87=88

Hence, the 30th term of sequence is 88.

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Write a rule for the function that shifts every point in the plane 10 units down and 4 units to the right
ElenaW [278]

Answer:

Substitute y with y+10 and x with x-4.

Step-by-step explanation:

To shift every point in the plane 10 down you have to substitute y with y+10 and to shift every point in the plane 4 units right you have to substitute x with x-4.

Suppose the origin of function on the plane is (0,0) now we have to make the origin (4,-10).

first (x,y)=(0,0)

When we substitute x and y

(x-4,y+10)=(0,0)

x-4=0    

x=4

y+10 = 0

y= -10

Now the origin is (4,-10).

Hence the function is shifted 10 units down and 4 units right.

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%7B%28%20%5Csqrt%7B3%7D%20%29%7D%5E%7Bx%20%2B%20y%7D%20%20%3D%209%20%5C%5C%20%20%7B%28%20%5
algol13

<h2>\bf \purple{ \underline{Given :-}}</h2>

• \:  {( \sqrt{3} )}^{x + y}  = 9\:  \:  \: \: (i)

• \:  {( \sqrt{2} )}^{x - y}  = 32 \:  \:  \: \: (ii)

\\

<h2>\bf \purple{ \underline{To  \: Find :- }}</h2>

•   {\sf{The  \: value  \: of}} \:   \: 2x+ y.

\\

\huge\bf \purple{ \underline{Solution :- }}

\sf{From \:  equation \:  (i), }

{( \sqrt{3} )}^{x + y}  = 9

⇒ {( \sqrt{3} )}^{x + y}  =  ({ \sqrt{3} })^{4}

⇒ x + y = 4 \:  \:  \: \: (iii)

\\

\sf{From \:  the  \: equation \:  (ii)}

{( \sqrt{2} )}^{x - y}  = 32

⇒( { \sqrt{2} })^{x - y}  =(  { \sqrt{2} })^{10}

⇒x - y = 10 \:  \:  \: \: (iv)

\\

\sf{We \:  have \:  to  \: add \:  equation \:  (iii)  \: and \:  equation  \: (iv)}

x + y + x - y = 4 + 10

⇒2x = 14

⇒x = 7

\\

\sf{Subtracting \:  equation \:  (iii)  \: from  \: equation  \: (ii),  }

x  - y - x - y = 10 - 4

⇒ - 2y = 6

⇒y =  - 3

\\

\bf\therefore x = 7 \: \:  and  \: \: y =  - 3

\\

{ \sf{Th e \:  value \:  of }}= 2x +y

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:   \: =2 \times 7  + ( - 3)

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:   \: =14 - 3

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:   =11

\bf \red{Hence,  \: the  \: value \:  of  \: 2x + y  \: is  \: 11. }

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Step-by-step explanation:

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