English, asked by nidhisingh01825, 6 months ago

9t2 - 3/2+ 1/16 when t is 2​

Answers

Answered by divyadharaasc
0

Answer:

Step by Step Solution:

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STEP

1

:

1

Simplify ——

16

Equation at the end of step

1

:

3 1

((9•(t2))+(—•t))+——

2 16

STEP

2

:

3

Simplify —

2

Equation at the end of step

2

:

3 1

((9 • (t2)) + (— • t)) + ——

2 16

STEP

3

:

Equation at the end of step

3

:

3t 1

(32t2 + ——) + ——

2 16

STEP

4

:

Rewriting the whole as an Equivalent Fraction

4.1 Adding a fraction to a whole

Rewrite the whole as a fraction using 2 as the denominator :

32t2 32t2 • 2

32t2 = ———— = ————————

1 2

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

4.2 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:

32t2 • 2 + 3t 18t2 + 3t

————————————— = —————————

2 2

Equation at the end of step

4

:

(18t2 + 3t) 1

——————————— + ——

2 16

STEP

5

:

STEP

6

:

Pulling out like terms

6.1 Pull out like factors :

18t2 + 3t = 3t • (6t + 1)

Calculating the Least Common Multiple :

6.2 Find the Least Common Multiple

The left denominator is : 2

The right denominator is : 16

Number of times each prime factor

appears in the factorization of:

Prime

Factor Left

Denominator Right

Denominator L.C.M = Max

{Left,Right}

2 1 4 4

Product of all

Prime Factors 2 16 16

Least Common Multiple:

16

Calculating Multipliers :

6.3 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 = 8

Right_M = L.C.M / R_Deno = 1

Making Equivalent Fractions :

6.4 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. 3t • (6t+1) • 8

—————————————————— = ———————————————

L.C.M 16

R. Mult. • R. Num. 1

—————————————————— = ——

L.C.M 16

Adding fractions that have a common denominator :

6.5 Adding up the two equivalent fractions

3t • (6t+1) • 8 + 1 144t2 + 24t + 1

——————————————————— = ———————————————

16 16

Trying to factor by splitting the middle term

6.6 Factoring 144t2 + 24t + 1

The first term is, 144t2 its coefficient is 144 .

The middle term is, +24t its coefficient is 24 .

The last term, "the constant", is +1

Step-1 : Multiply the coefficient of the first term by the constant 144 • 1 = 144

Step-2 : Find two factors of 144 whose sum equals the coefficient of the middle term, which is 24 .

-144 + -1 = -145

-72 + -2 = -74

-48 + -3 = -51

-36 + -4 = -40

-24 + -6 = -30

-18 + -8 = -26

-16 + -9 = -25

-12 + -12 = -24

-9 + -16 = -25

-8 + -18 = -26

-6 + -24 = -30

-4 + -36 = -40

-3 + -48 = -51

-2 + -72 = -74

-1 + -144 = -145

1 + 144 = 145

2 + 72 = 74

3 + 48 = 51

4 + 36 = 40

6 + 24 = 30

8 + 18 = 26

9 + 16 = 25

12 + 12 = 24 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 12 and 12

144t2 + 12t + 12t + 1

Step-4 : Add up the first 2 terms, pulling out like factors :

12t • (12t+1)

Add up the last 2 terms, pulling out common factors :

1 • (12t+1)

Step-5 : Add up the four terms of step 4 :

(12t+1) • (12t+1)

Which is the desired factorization

Multiplying Exponential Expressions:

6.7 Multiply (12t+1) by (12t+1)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (12t+1) and the exponents are :

1 , as (12t+1) is the same number as (12t+1)1

and 1 , as (12t+1) is the same number as (12t+1)1

The product is therefore, (12t+1)(1+1) = (12t+1)2

Final result :

(12t + 1)2

——————————

16

Answered by Anonymous
6

Answer:

hope it helps u mate

iTz sAnA _______

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