Work out the value of 3 \times(3^2 + 4) - 83×(32+4)−8

Conversion a mixed number 8 1/

4

to a improper fraction: 8 1/4 = 8 1/

4

= 8 · 4 + 1/

4

= 32 + 1/

4

= 33/

4

To find new numerator:

a) Multiply the whole number 8 by the denominator 4. Whole number 8 equally 8 * 4/

4

= 32/

4

b) Add the answer from previous step 32 to the numerator 1. New numerator is 32 + 1 = 33

c) Write a previous answer (new numerator 33) over the denominator 4.

Eight and one quarter is thirty-three quarters

Conversion a mixed number 3 2/

5

to a improper fraction: 3 2/5 = 3 2/

5

= 3 · 5 + 2/

5

= 15 + 2/

5

= 17/

5

To find new numerator:

a) Multiply the whole number 3 by the denominator 5. Whole number 3 equally 3 * 5/

5

= 15/

5

b) Add the answer from previous step 15 to the numerator 2. New numerator is 15 + 2 = 17

c) Write a previous answer (new numerator 17) over the denominator 5.

Three and two fifths is seventeen fifths

Subtract: 33/

4

- 17/

5

= 33 · 5/

4 · 5

- 17 · 4/

5 · 4

= 165/

20

- 68/

20

= 165 - 68/

20

= 97/

20

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(4, 5) = 20. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 5 = 20. In the next intermediate step the fraction result cannot be further simplified by canceling.

In words - thirty-three quarters minus seventeen fifths = ninety-seven twentieths.

Conversion a mixed number 2 1/

3

to a improper fraction: 2 1/3 = 2 1/

3

= 2 · 3 + 1/

3

= 6 + 1/

3

= 7/

3

To find new numerator:

a) Multiply the whole number 2 by the denominator 3. Whole number 2 equally 2 * 3/

3

= 6/

3

b) Add the answer from previous step 6 to the numerator 1. New numerator is 6 + 1 = 7

c) Write a previous answer (new numerator 7) over the denominator 3.

Two and one third is seven thirds

Subtract: 7/

3

- 1/

4

= 7 · 4/

3 · 4

- 1 · 3/

4 · 3

= 28/

12

- 3/

12

= 28 - 3/

12

= 25/

12

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(3, 4) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 3 × 4 = 12. In the next intermediate step the fraction result cannot be further simplified by canceling.

In words - seven thirds minus one quarter = twenty-five twelfths.

Subtract: the result of step No. 3 - the result of step No. 5 = 97/

20

- 25/

12

= 97 · 3/

20 · 3

- 25 · 5/

12 · 5

= 291/

60

- 125/

60

= 291 - 125/

60

= 166/

60

= 2 · 83/

2 · 30

= 83/

30

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(20, 12) = 60. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 20 × 12 = 240. In the next intermediate step , cancel by a common factor of 2 gives 83/

30

.

In words - ninety-seven twentieths minus twenty-five twelfths = eighty-three thirtieths.

Step-by-step explanation:took a lot of time brainliest plz

the awnser is visionary

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.