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vladimir2022 [97]
4 years ago
6

A square has of 30 square meters and a perimeter of 34 meters. what ate the dimensions of the rectangle?

Mathematics
1 answer:
USPshnik [31]4 years ago
3 0
34 = 2b+2h
17 = b+h
b = 17-h

30 = b*h
30 = (17-h)h
30 = 17h - h^2
h^2 - 17h + 30 = 0
(h - 15)(h - 2) = 0
h = 2 or h = 15
if h = 2 then b = 30/2 = 15
if h = 15 then b = 30/15 = 2

so one side is 15 long and one side is 2 long
by the way its not a square
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3 years ago
In an arithmetic sequence, a17 = -40 and
Viktor [21]

Answer:

Tn = 2Tn-1 - Tn-2

Step-by-step explanation:

Before we can generate the recursive sequence, we need to find the nth term of the given sequence.

nth term of an AP is given as:

Tn = a+(n-1)d

If a17 = -40

T17 = a+(17-1)d = -40

a+16d = -40 ...(1)

If a28 = -73

T28 = a+(28-1)d = -73

a+27d = -73 ...(2)

Solving both equations simultaneously using elimination method.

Subtracting 1 from 2 we have:

27d - 16d = -73-(-40)

11d = -73+40

11d = -33

d = -3

Substituting d = -3 into 1

a+16(-3) = -40

a - 48 = -40

a = -40+48

a = 8

Given a = 8, d = -3, the nth term of the sequence will be

Tn = 8+(n-1) (-3)

Tn = 8+(-3n+3)

Tn = 8-3n+3

Tn = 11-3n

Given Tn = 11-3n and d = -3

Tn-1 = Tn - d... (3)

Tn-1 = 11-3n +3

Tn-1 = 14-3n

Tn-2 = Tn-2d...(4)

Tn-2 = 11-3n-2(-3)

Tn-2 = 11-3n+6

Tn-2 = 17-3n

From 3, d = Tn - Tn-1

From 4, d = (Tn - Tn-2)/2

Equating both common difference

(Tn - Tn-2)/2 = Tn - Tn-1

Tn - Tn-2 = 2(Tn - Tn-1)

Tn - Tn-2 = 2Tn-2Tn-1

2Tn-Tn = 2Tn-1 - Tn-2

Tn = 2Tn-1 - Tn-2

The recursive formula will be

Tn = 2Tn-1 - Tn-2

5 0
3 years ago
Read 2 more answers
NEED HELP FAST!!!!!
slava [35]
5p-4p-8=-2+3

Combine like terms
p-8= 1

Add 8 to both sides to isolate p
p=9

Final answer: 1 solution only, p=9
8 0
3 years ago
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