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3 years ago

given that (y+2) (y+3) and (2y^2+1) are consecutive terms of an arithmetic progression, find the possible value of y

Mathematics

1 answer:

mafiozo [28]3 years ago

8 0

Answer:

Step-by-step explanation:

Given the arithmetic sequence (y+2) (y+3) and (2y²+1), the common difference is gotten by taking the difference in their terms. For example if we have 3 terms T1, T2, T3... the common difference d = T2-T1 = T3-T2

From the sequence given;

T1 = y+2, T2 = y+3 and T3 = 2y²+1

d = y+3-(y+2) = 2y²+1- (y+3)

open the parenthesis

y+3-y-2 = 2y²+1- y-3

1 = 2y²+1- y-3

1 = 2y²- y-2

2y²- y-2-1 = 0

2y²- y-3 =0

Factorize the resulting expression

2y²- y-3 =0

2y²- 2y+3y-3 =0

2y(y-1)+3(y-1) = 0

(2y+3)(y-1) = 0

2y+3 = 0 and y-1 = 0

2y = -3 and y =1

y = -3/2 and 1

<em>Hence the possible values of y are -3/2 and 1</em>

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1 year ago

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hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

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3 years ago

given that (y+2) (y+3) and (2y^2+1) are consecutive terms of an arithmetic progression, find the possible value of y​

given that (y+2) (y+3) and (2y^2+1) are consecutive terms of an arithmetic progression, find the possible value of y