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aleksandr82 [10.1K]
3 years ago
10

If x = 3 cm and y = 2 cm, what is the perimeter of the polygon?

Mathematics
2 answers:
lara [203]3 years ago
6 0
10! This is the correct answer for apex
Temka [501]3 years ago
3 0
Answer: P=(3*2)+(2*2)=10cm
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9(2m + 5n + 10)<br> Distributive Property
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18m + 45n + 90

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A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20
Alexxandr [17]

Answer:

The smallest possible value for the third term of the geometric progression is 1.

Step-by-step explanation:

Given : A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression.

To find : What is the smallest possible value for the third term of the geometric progression?  

Solution :

The arithmetic progression is given by a,a+d,a+2d

First term a=9

Second term - 2 is added to second term i.e. a+d+2=9+d+2=11+d

Third term - 20 is added to the third term i.e.a+2d+20=9+2d+20=29+2d

The geometric progression is given by a,ar,ar^2

First term a=9

Second term -ar=11+d

Third term ar^2=29+2d

r is the common ratio which is second term divided by first term,

So, r=\frac{ar}{a}=\frac{11+d}{9}

or  third term divided by second term,

So, r=\frac{ar^2}{ar}=\frac{29+2d}{11+d}

Equating both the r,

\frac{11+d}{9}=\frac{29+2d}{11+d}

Cross multiply,

(11+d)\times (11+d)=(29+2d)(9)

11^2+11d+11d+d^2=261+18d

121+22d+d^2-261-18d=0

d^2+4d-140=0

Solving by middle term split,

d^2+14d-10d-140=0

d(d+14)-10(d+14)=0

(d+14)(d-10)=0

d=-14,10

Substituting the value of d in the third term,

Third term ar^2=29+2d

When d=-14,

Third term ar^2=29+2(-14)=29-28=1

When d=10,

Third term ar^2=29+2(10)=29+20=49

Therefore, The smallest possible value for the third term of the geometric progression is 1.

7 0
3 years ago
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