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olga nikolaevna [1]
3 years ago
10

Find common ratio and series sum

Mathematics
1 answer:
Artyom0805 [142]3 years ago
6 0

Answer:

Her total harvest over the

first 5 years that she owns the farm is 11277 kg

Step-by-step explanation:

Maggie is starting a small farm. In her first year, she harvests 90.000 kilograms of kale.

As she gets better at planting, fertilizing and pest control,

each year her harvest increases by 3%.

So,

In First year

She owns 90,000 kg

In Second year

She owns 90,000 kg and her harvest is increases by 3%.

i.e.,

90000 × 3 ÷ 100 = 2700

So, The increase in 90,000 kg kale is 2,700

Now,

90,000 + 2,700 = 92,700 kg

In Third year

She owns 92,700 kg and her harvest is increases by 3%.

i.e.,

92700 × 3 ÷ 100 = 2781

So, The increase in 92,700 kg kale is 2781

Now,

92,700 + 2,781 = 95,481 kg

In Fourth year

She owns 95,481 kg and her harvest is increases by 3%.

i.e.,

95481 × 3 ÷ 100 = 2846

So, The increase in 95,481 kg kale is 2846

Now,

95,481 + 2,846 = 98,345 kg

In Fifth year

She owns 98,354 kg and her harvest is increases by 3%.

i.e.,

98354 × 3 ÷ 100 = 2950

So, The increase in 98354 kg kale is 2950

Now,

98354 + 2950 = 101,295 kg

Now,

In First five years She owns

2700 + 2781 + 2846 + 2950 = 11277 kg

Thus, her total harvest over the first 5 years that she owns the farm is 11277 kg

<u>-TheUnknownScientist</u>

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leva [86]

Final result :

(b - a) • (a2 + ab + b2)

————————————————————————

a2b3

Step by step solution :

Step 1 :

1

Simplify —

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Equation at the end of step 1 :

1 1 1

————-———— ÷ (—•b)

(a2) (b2) a

Step 2 :

1

Simplify ——

b2

Equation at the end of step 2 :

1 1 b

———— - —— ÷ —

(a2) b2 a

Step 3 :

1 b

Divide —— by —

b2 a

3.1 Dividing fractions

To divide fractions, write the divison as multiplication by the reciprocal of the divisor :

1 b 1 a

—— ÷ — = —— • —

b2 a b2 b

Multiplying exponential expressions :

3.2 b2 multiplied by b1 = b(2 + 1) = b3

Equation at the end of step 3 :

1 a

———— - ——

(a2) b3

Step 4 :

1

Simplify ——

a2

Equation at the end of step 4 :

1 a

—— - ——

a2 b3

Step 5 :

Calculating the Least Common Multiple :

5.1 Find the Least Common Multiple

The left denominator is : a2

The right denominator is : b3

Number of times each Algebraic Factor

appears in the factorization of:

Algebraic

Factor Left

Denominator Right

Denominator L.C.M = Max

{Left,Right}

a 2 0 2

b 0 3 3

Least Common Multiple:

a2b3

Calculating Multipliers :

5.2 Calculate multipliers for the two fractions

Denote the Least Common Multiple by L.C.M

Denote the Left Multiplier by Left_M

Denote the Right Multiplier by Right_M

Denote the Left Deniminator by L_Deno

Denote the Right Multiplier by R_Deno

Left_M = L.C.M / L_Deno = b3

Right_M = L.C.M / R_Deno = a2

Making Equivalent Fractions :

5.3 Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. b3

—————————————————— = ————

L.C.M a2b3

R. Mult. • R. Num. a • a2

—————————————————— = ——————

L.C.M a2b3

Adding fractions that have a common denominator :

5.4 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

b3 - (a • a2) b3 - a3

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a2b3 a2b3

Trying to factor as a Difference of Cubes:

5.5 Factoring: b3 - a3

Theory : A difference of two perfect cubes, a3 - b3 can be factored into

(a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =

a3+a2b+ab2-ba2-b2a-b3 =

a3+(a2b-ba2)+(ab2-b2a)-b3 =

a3+0+0+b3 =

a3+b3

Check : b3 is the cube of b1

Check : a3 is the cube of a1

Factorization is :

(b - a) • (b2 + ab + a2)

Trying to factor a multi variable polynomial :

5.6 Factoring b2 + ab + a2

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Final result :

(b - a) • (a2 + ab + b2)

————————————————————————

a2b3

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boyakko [2]

Explanation:

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{3 \choose 1} = \frac{3!}{1!(3-1)!} = \frac{6}{1*2} = 3

b) First, we have that

{n \choose k} = \frac{n!}{k!(n-k)!}

On the other hand,

{n \choose n-k} = \frac{n!}{(n-k)!(n(n-k))!} = \frac{n!}{(n-k)!k!}

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