Simplify: -(16 - 5x) 5×-16/ 5× + 16 / -5×-16/ -5 + 16

(16 (1280 x - 4071))/25

Step-by-step explanation:

Simplify the following:

16 - (-16 (-16)×5×16 (16 - 5 x))/(5 (-5) (-5))

Hint: | Express 16/(-5)×(-16)/(-5) as a single fraction.

16/(-5)×(-16)/(-5) = (16 (-16))/(-5 (-5)):

-(-16×5×(-16×16)/(-5 (-5)) (16 - 5 x))/5 + 16

Hint: | Express 5×(-16)/5×(16 (-16))/(-5 (-5)) as a single fraction.

5×(-16)/5×(16 (-16))/(-5 (-5)) = (16 (-16) 5 (-16))/(-5 (-5) 5):

16 - (-16 (-16) 16×5)/(-5 (-5) 5) (16 - 5 x)

Hint: | Multiply 16 and -16 together.

16 (-16) = -256:

-(-16-256×5 (16 - 5 x))/(-5 (-5) 5) + 16

Hint: | Multiply -256 and 5 together.

-256×5 = -1280:

-(-16-1280 (16 - 5 x))/(-5 (-5) 5) + 16

Hint: | Multiply -1280 and -16 together.

-1280 (-16) = 20480:

16 - (20480 (16 - 5 x))/(-5 (-5) 5)

Hint: | In 20480/(-5 (-5) 5), divide 20480 in the numerator by 5 in the denominator.

5 | | 4 | 0 | 9 | 6

| 2 | 0 | 4 | 8 | 0

- | 2 | 0 | | |  

| | | 4 | |  

| | - | 0 | |  

| | | 4 | 8 |  

| | - | 4 | 5 |  

| | | | 3 | 0

| | | - | 3 | 0

| | | | | 0:

16 - (4096 (16 - 5 x))/(-5 (-5))

Hint: | Multiply -5 and -5 together.

-5 (-5) = 25:

16 - (4096 (16 - 5 x))/25

Hint: | Put the fractions in 16 - ((16 - 5 x)×4096)/25 over a common denominator.

Put each term in 16 - ((16 - 5 x)×4096)/25 over the common denominator 25: 16 - ((16 - 5 x)×4096)/25 = 400/25 - (4096 (16 - 5 x))/25:

400/25 - (4096 (16 - 5 x))/25

Hint: | Combine 400/25 - (4096 (16 - 5 x))/25 into a single fraction.

400/25 - (4096 (16 - 5 x))/25 = (400 - 4096 (16 - 5 x))/25:

(400 - 4096 (16 - 5 x))/25

Hint: | Distribute -4096 over 16 - 5 x.

-4096 (16 - 5 x) = 20480 x - 65536:

(20480 x - 65536 + 400)/25

Hint: | Group like terms in 20480 x - 65536 + 400.

Grouping like terms, 20480 x - 65536 + 400 = 20480 x + (400 - 65536):

(20480 x + (400 - 65536))/25

Hint: | Evaluate 400 - 65536.

400 - 65536 = -65136:

(20480 x + -65136)/25

Hint: | Factor out the greatest common divisor of the coefficients of 20480 x - 65136.

Factor 16 out of 20480 x - 65136:

(16 (1280 x - 4071))/25

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐