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Eddi Din [679]
3 years ago
14

Given the third term of an arithmetic sequence less than the fourth term by three. The seventh term

Mathematics
1 answer:
UNO [17]3 years ago
6 0

Answer:

The first term is -6 and common difference is 3

Step-by-step explanation:

The nth term of an arithmetic progression is expressed as;

Tn = a + (n-1)d

T3 = a+2d

T4 = a + 3d

If the third term of an arithmetic sequence less than the fourth term by three

T3 = T4 - 3

a+2d = a+3d - 3

2d - 3d = -3

-d  = -3

d = 3

Similarly;

T7 = a+6d

T5 = a+4d

If the seventh term  is two times the fifth term, the;

T7 = 2T5

a+6d = 2(a+4d)

a+6d = 2a+8d

Since d = 3

a + 6(3) = 2a + 8(3)

a+18 = 2a + 24

a-2a = 24 - 18

-a = 6

a = -6

Hence the first term is -6 and common difference is 3

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ZanzabumX [31]
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The answer is 4 1/2.

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3 years ago
Question: A square with sides of length 5 is positioned inside a square with sides of length 7. What is the total area between t
Sergio [31]

Answer:

The total area between the squares is of 24 units squared.

Step-by-step explanation:

Area of a square:

The area of a square of side length l is given by:

A = l^2

Area between figures:

The area between two figures is the area of the larger figure subtracted by the area of the smaller figure.

Larger square:

Square with side of length 7. So

A_l = 7^2 = 49

Smaller square:

Square with side of length 5. So

A_s = 5^2 = 25

What is the total area between the squares? ​

T = A_l - A_s = 49 - 25 = 24

The total area between the squares is of 24 units squared.

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5 0
3 years ago
Find the term that must be added to the equation x2 6x=1 to make it into a perfect square.
andrezito [222]

The value added to the equation $x^2-6x=1 exists $x^2-6x+9=10.

<h3>What is a perfect square?</h3>

A perfect square exists as a number that can be described as the product of an integer by itself or as the second exponent of an integer.

The perfect square trinomial exists

(a ± b)^2 = a ^2 ± 2ab + b ^2

$x^2-6x=1

x^2-2*3x=1

then $2ab = 2 * 3*x = 2 * x *3

The value of a = x and b = 3

$b^2=3^2=9

$x^2-6x+9=1+9

$x^2-6x+9=10

The value added to the equation $x^2-6x=1 exists $x^2-6x+9=10.

To learn more about perfect square refer to: brainly.com/question/6946048

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3 0
2 years ago
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