Transcript
Page 1: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

Resoluรงรฃo โ€“ Meta 6

Page 2: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

A) x = 1,5

f(1,5) = -3.1,5 + 4

f(1,5) = -0,5

B) x = a + b

f(a + b) = -3.(a + b) + 4

f(a + b) = -3a - 3b + 4

QUESTรƒO 1.

Page 3: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

a) Constante

b) Linear

c) Afim

d) Afim

QUESTรƒO 2.

Page 4: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

f(x) = -2x + 3

a)f(1) = -2.1 + 3 = 1

b)f(0) = -2.0 + 3 = 3

c) f(๐Ÿ

๐Ÿ‘) = -2 .

๐Ÿ

๐Ÿ‘+ 3 =

๐Ÿ•

๐Ÿ‘

d) f(-๐Ÿ

๐Ÿ) = -2.

โˆ’๐Ÿ

๐Ÿ+ 3 = 4

QUESTรƒO 3.

Page 5: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

a)f(x) = 1

2x + 3 = 1

x = -1

b)f(x) = 0

0 = 2x + 3

x = -3/2

QUESTรƒO 4.

c) f(x) = ๐Ÿ

๐Ÿ‘๐Ÿ

๐Ÿ‘= 2x + 3 (.3)

1 = 6x + 9x = -8/6

d) f(x) = 0,750,75 = 2x + 3๐Ÿ‘

๐Ÿ’= 2x + 3 (.4)

3 = 8x + 12x = -9/8

Page 6: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 5.

a) f(x) = 4x + 5

Taxa de variaรงรฃo = 4

b) f(x) = 3

Taxa de variaรงรฃo = 0

Page 7: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 6.

f(x + h) โ€“ f(x).

a) h(x) =๐Ÿ

๐Ÿ๐’™ + ๐Ÿ‘

=๐Ÿ .(๐’™ + ๐’‰)

๐Ÿ+ ๐Ÿ‘ โ€“ (

๐Ÿ

๐Ÿ๐’™ + ๐Ÿ‘)

=๐’™

๐Ÿ+

๐’‰

๐Ÿ+ 3 -

๐’™

๐Ÿโˆ’ ๐Ÿ‘

= ๐’‰

๐Ÿ

Afim, pois nรฃo depende de x.

b) f(x) = 2xยฒ

= 2 . (x + h) - 2xยฒ

= 2x + 2h - 2xยฒ

Nรฃo รฉ afim, pois depende de x.

Page 8: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 7.

A)

x

y

-2

-1

B)

x

y

-1

3

Page 9: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 8.

A)

x

y

0

B)

C)

x

y

5

F)

x

y

-3

D)

x

y

3

E)

x

y

0

x

y

0 -3

Page 10: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 9.

x

y

f

g

h

s

t

Page 11: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 10.

a) R: 4x + 3y + 7 = 0 e A(2,3)

d =4 . 2 + 3 . 3 + 7

๐Ÿ’ ๐Ÿ+ ๐Ÿ‘ ๐Ÿ=๐Ÿ๐Ÿ’

๐Ÿ“

b) s: 3x โ€“ 4y + 2 = 0 e A(-1,2)

d = 3.โˆ’1 โˆ’ 4.2 + 2

๐Ÿ‘ ๐Ÿ+ โˆ’๐Ÿ’ ๐Ÿ= ๐Ÿ—

๐Ÿ“

c) t: 12x + 13y + 8 = 0 e A(2, 2)

d =12.2 + 13.2 + 8

๐Ÿ๐Ÿ ๐Ÿ+ ๐Ÿ๐Ÿ‘ ๐Ÿ=

๐Ÿ“๐Ÿ–

๐Ÿ‘๐Ÿ๐Ÿ‘

d) z: 5x โ€“ 4y + 8 = 0 e A(1, - 2)

d =5.1 โˆ’ 4.โˆ’2 + 8

๐Ÿ“ ๐Ÿ+ โˆ’๐Ÿ’ ๐Ÿ=

๐Ÿ๐Ÿ

๐Ÿ’๐Ÿ

e) v: 3x + 4y + 8 = 0 e A(2,1)

d =3.2 + 4.1 + 8

๐Ÿ‘ ๐Ÿ+ ๐Ÿ’ ๐Ÿ=๐Ÿ๐Ÿ–

๐Ÿ“

Page 12: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 11.

a) f(x)= - 2x

Com eixo x:

โˆ’๐ŸŽ

โˆ’๐Ÿ= ๐ŸŽ

Com eixo y:

0

b) f(x) =๐Ÿ

๐Ÿ๐’™ โˆ’ ๐Ÿ

Com eixo x:

โˆ’(โˆ’๐Ÿ)

๐Ÿ๐Ÿ

= ๐Ÿ

Com eixo y:

-1

Page 13: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 12.

a) Afim

b) Afim

c) Constante

d) identidade

e) Linear

f) Translaรงรฃo

Page 14: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 13.

a) f(x) = 3(x+1) + 4(x โ€“ 1)

f(x) = 3x + 3 + 4x โ€“ 4

f(x) = 7x โ€“ 1

a = 7 e b = -1

b) f(x) = (x+2)ยฒ + (x+2)(x-2)

f(x) = ๐’™๐Ÿ + ๐Ÿ’๐ฑ + ๐Ÿ’ + ๐’™๐Ÿ - 4

f(x) = 2๐’™๐Ÿ + 4x

Nรฃo รฉ funรงรฃo afim.

c) f(x) = (x-3)ยฒ - x(x-5)

f(x) = xยฒ - 6x + 9 - xยฒ + 5x

f(x) = -x + 9

a = -1 e b = 9

d) f(x) = (x-3) โ€“ 5(x-1)

f(x) = x โ€“ 3 โ€“ 5x + 5

f(x) = -4x + 2

a = -4 e b = 2

Page 15: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 14.

A)

x

y5

1-3

-7

B)

x

y7

2-1

1

Page 16: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 15.

Taxa de variaรงรฃo = -3

Page 17: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 16.a) A(1, 2) e B(3, 4)

d = ๐Ÿ โˆ’ ๐Ÿ‘ ๐Ÿ + ๐Ÿ โˆ’ ๐Ÿ’ ๐Ÿ

d = ๐Ÿ–

๐‘ท๐‘ด๐’™ =๐Ÿ+๐Ÿ‘

๐Ÿ= 2

๐‘ท๐‘ด๐’š =๐Ÿ+๐Ÿ’

๐Ÿ= 3 ๐‘ท๐‘ด(๐Ÿ; ๐Ÿ‘)

b) A(-1, 0) e B(2, 7)

d = โˆ’๐Ÿ โˆ’ ๐Ÿ ๐Ÿ + ๐ŸŽ โˆ’ ๐Ÿ• ๐Ÿ

d = ๐Ÿ“๐Ÿ–

๐‘ท๐‘ด๐’™ =โˆ’๐Ÿ+๐Ÿ

๐Ÿ= 0,5

๐‘ท๐‘ด๐’š =๐ŸŽ + ๐Ÿ•

๐Ÿ= 3,5 ๐‘ท๐‘ด(๐ŸŽ, ๐Ÿ“; ๐Ÿ‘, ๐Ÿ“)

c) A(2, 9) e D(4, -5)

d = ๐Ÿ โˆ’ ๐Ÿ’ ๐Ÿ + ๐Ÿ— โˆ’ (โˆ’๐Ÿ“) ๐Ÿ

d = ๐Ÿ๐ŸŽ๐ŸŽ

๐‘ท๐‘ด๐’™ =๐Ÿ + ๐Ÿ’

๐Ÿ= 3

๐‘ท๐‘ด๐’š =๐Ÿ— + (โˆ’๐Ÿ“)

๐Ÿ= 2 ๐‘ท๐‘ด(๐Ÿ‘; ๐Ÿ)

d) A(3, 4) e E(2, -1)

d = ๐Ÿ‘ โˆ’ ๐Ÿ ๐Ÿ + ๐Ÿ’ โˆ’ (โˆ’๐Ÿ) ๐Ÿ

d = ๐Ÿ๐Ÿ”

๐‘ท๐‘ด๐’™ =๐Ÿ‘ +๐Ÿ

๐Ÿ= 2,5

๐‘ท๐‘ด๐’š =๐Ÿ’ + (โˆ’๐Ÿ)

๐Ÿ= 1,5 ๐‘ท๐‘ด(๐Ÿ, ๐Ÿ“; ๐Ÿ, ๐Ÿ“)

Page 18: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 17.

a) A(2, 4) e B(6, 2)

Coeficiente Angular = โˆ†๐ฒ

โˆ†๐ฑ=

๐Ÿ’ โˆ’ ๐Ÿ

๐Ÿ โˆ’๐Ÿ”=

๐Ÿ

โˆ’๐Ÿ’= -

๐Ÿ

๐Ÿ

Como as retas sรฃo perpendiculares: โˆ’๐Ÿ

๐Ÿ. ๐ฆ๐ฌ = โˆ’๐Ÿ

๐ฆ๐ฌ = ๐Ÿ

๐‘ท๐‘ด๐’™ =๐Ÿ+๐Ÿ”

๐Ÿ= 4

๐‘ท๐‘ด๐’š =๐Ÿ’ +๐Ÿ

๐Ÿ= 3 ๐‘ท๐‘ด(๐Ÿ’; ๐Ÿ‘)

y = ax + b3 = 2.4 + bb = -5

y = 2x - 5

Page 19: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 17.

b) A(-3, 1) e B(1, 5)

Coeficiente Angular = โˆ†๐ฒ

โˆ†๐ฑ=

๐Ÿ โˆ’ ๐Ÿ“

โˆ’๐Ÿ‘ โˆ’ ๐Ÿ=

โˆ’๐Ÿ’

โˆ’๐Ÿ’= 1

Como as retas sรฃo perpendiculares: 1 . ๐ฆ๐ฌ = โˆ’๐Ÿ๐ฆ๐ฌ = โˆ’๐Ÿ

๐‘ท๐‘ด๐’™ =โˆ’๐Ÿ‘ +๐Ÿ

๐Ÿ= -1

๐‘ท๐‘ด๐’š =๐Ÿ +๐Ÿ“

๐Ÿ= 3 ๐‘ท๐‘ด(โˆ’๐Ÿ; ๐Ÿ‘)

y = ax + b3 = -1 . -1 + bb = 2 y = -1x + 2

Page 20: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 18.a) (r): 2x โ€“ 3y + 7 = 0 e P(2, 3)

2x โ€“ 3y + 7 = 0

y =2๐‘ฅ

3+

7

3

Como as retas sรฃo perpendiculares: 2

3. ๐ฆ๐ฌ = โˆ’๐Ÿ

๐ฆ๐ฌ =โˆ’3

2y = ax + b

3 = โˆ’3

2. 2 + b

b = 6

b) (r): 4x โ€“ 3y + 1 = 0 e P(0, 0)

4x โ€“ 3y + 1 = 0

y =๐Ÿ’๐’™

๐Ÿ‘+

๐Ÿ

๐Ÿ‘

Como as retas sรฃo perpendiculares: ๐Ÿ’

๐Ÿ‘. ๐ฆ๐ฌ = โˆ’๐Ÿ

๐ฆ๐ฌ =โˆ’๐Ÿ‘

๐Ÿ’y = ax + b

0 = โˆ’๐Ÿ‘

๐Ÿ’. ๐ŸŽ + b

b = 0y = โˆ’3๐ฑ

2+ 6 y =

โˆ’3๐ฑ

๐Ÿ’

Page 21: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 18.c) (r): y = 3x โ€“ 2 e P(3, -3)

Como as retas sรฃo perpendiculares: 3 . ๐ฆ๐ฌ = โˆ’๐Ÿ

๐ฆ๐ฌ =โˆ’๐Ÿ

๐Ÿ‘y = ax + b

-3 = โˆ’๐Ÿ

๐Ÿ‘. ๐Ÿ‘ + b

b = -2 y = โˆ’๐ฑ

๐Ÿ‘- 2

Page 22: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 19.a) A(1, 2) e B(3, 4)

Coeficiente Angular = โˆ†๐’š

โˆ†๐’™=

๐Ÿ โˆ’๐Ÿ’

๐Ÿ โˆ’ ๐Ÿ‘=

โˆ’๐Ÿ

โˆ’๐Ÿ= ๐Ÿ

y = ax + b2 = 1.1 + bb = 1

y = 1x + 1

Page 23: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 19.b) A(-1, 0) e B(2, 7)

Coeficiente Angular = โˆ†๐’š

โˆ†๐’™=

๐ŸŽ โˆ’๐Ÿ•

โˆ’๐Ÿ โˆ’๐Ÿ=

โˆ’๐Ÿ•

โˆ’๐Ÿ‘=

๐Ÿ•

๐Ÿ‘

y = ax + b

0 = ๐Ÿ•

๐Ÿ‘. -1 + b

b = ๐Ÿ•

๐Ÿ‘y =

๐Ÿ•

๐Ÿ‘x +

๐Ÿ•

๐Ÿ‘

Page 24: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 19.c) A(2, 9) e D(4, -5)

Coeficiente Angular = โˆ†๐’š

โˆ†๐’™=

๐Ÿ— โˆ’(โˆ’๐Ÿ“)

๐Ÿ โˆ’ ๐Ÿ’=

๐Ÿ๐Ÿ’

โˆ’๐Ÿ= -7

y = ax + b9 = -7. 2 + bb = 23

y = -7x + 23

Page 25: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 19.d) A(3, 4) e E(2, -1)

Coeficiente Angular = โˆ†๐’š

โˆ†๐’™=

๐Ÿ’ โˆ’(โˆ’๐Ÿ)

๐Ÿ‘ โˆ’๐Ÿ=

๐Ÿ“

๐Ÿ= 5

y = ax + b4 = 5.3 + bb = -11 y = 5x - 11

Page 26: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 20.

a) 4x + 3y + 7 = 0

y =โˆ’๐Ÿ’๐’™

๐Ÿ‘-๐Ÿ•

๐Ÿ‘

Coef. Angular =โˆ’๐Ÿ’

๐Ÿ‘

b) -5x โ€“ 7y + 9 = 0

y =โˆ’๐Ÿ“๐’™

๐Ÿ•+๐Ÿ—

๐Ÿ•

Coef. Angular =โˆ’๐Ÿ“

๐Ÿ•

c) 3x + 7y โ€“ 9 = 0

y =โˆ’๐Ÿ‘๐’™

๐Ÿ•+๐Ÿ•

๐Ÿ‘

Coef. Angular =โˆ’๐Ÿ‘

๐Ÿ•

d) 5x โ€“ 4y + 8 = 0

y =๐Ÿ“๐’™

๐Ÿ’+๐Ÿ–

๐Ÿ’

Coef. Angular =๐Ÿ“

๐Ÿ’

e) 7x + 6y + 9 = 0

y =โˆ’๐Ÿ•๐’™

๐Ÿ”-๐Ÿ—

๐Ÿ”

Coef. Angular =โˆ’๐Ÿ•

๐Ÿ”

f)A(1, 2) e B(3, 4)๐Ÿ โˆ’๐Ÿ’

๐Ÿ โˆ’๐Ÿ‘=โˆ’๐Ÿ

โˆ’๐Ÿ=1

Coef. Angular = 1

g)A(-1, 0) e B(2, 7)๐ŸŽ โˆ’๐Ÿ•

โˆ’๐Ÿ โˆ’ ๐Ÿ=โˆ’๐Ÿ•

โˆ’๐Ÿ‘=

๐Ÿ•

๐Ÿ‘

Coef. Angular =๐Ÿ•

๐Ÿ‘

h)A(2, 9) e D(4, -5)๐Ÿ— โˆ’(โˆ’๐Ÿ“)

๐Ÿ โˆ’๐Ÿ’=๐Ÿ๐Ÿ’

โˆ’๐Ÿ= -7

Coef. Angular = -7

i)A(3, 4) e E(2, -1)๐Ÿ’ โˆ’(โˆ’๐Ÿ)

๐Ÿ‘ โˆ’ ๐Ÿ=๐Ÿ“

๐Ÿ= ๐Ÿ“

Coef. Angular = 5

Page 27: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 21.

a) = 45ยฐCoeficiente Angular = tg 45ยฐ= 1

b) = 120ยฐCoeficiente Angular = tg 120ยฐ= - tg 60ยฐ= โˆ’ ๐Ÿ‘

c) = 150ยฐ

Coeficiente Angular = tg 150ยฐ = - tg 30ยฐ= -๐Ÿ‘

๐Ÿ‘

d) = 60ยฐCoeficiente Angular = tg 60ยฐ= ๐Ÿ‘

e) = 135ยฐCoeficiente Angular = tg 135ยฐ = - tg 45ยฐ= -1

Page 28: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 22.a) = 45ยฐ e P(1, 2)Coeficiente Angular = tg 45ยฐ= 1

b) = 30ยฐ e P(2, 2)

Coeficiente Angular = tg 30ยฐ=๐Ÿ‘

๐Ÿ‘

c) = 120ยบ e P(-1, 6)

Coeficiente Angular = tg 120ยฐ= - tg 60ยฐ= โˆ’ ๐Ÿ‘

y = ax + b

6 = โˆ’ ๐Ÿ‘.-1 + b

b = 6 โˆ’ ๐Ÿ‘

y = โˆ’ ๐Ÿ‘ x + 6 โˆ’ ๐Ÿ‘

y = ax + b

2 = ๐Ÿ‘

๐Ÿ‘.2 + b

b = 2 -๐Ÿ ๐Ÿ‘

๐Ÿ‘

y = ๐Ÿ‘

๐Ÿ‘x + 2 -

๐Ÿ ๐Ÿ‘

๐Ÿ‘

y = ax + b2 = 1.1 + bb = 1

y = 1x + 1

Page 29: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 22.

d) = 60ยบ e P(8, 0)

Coeficiente Angular = tg 60ยฐ= ๐Ÿ‘

e) = 150ยบ e P(6, -2)

Coeficiente Angular = tg 150ยฐ = - tg 30ยฐ= -๐Ÿ‘

๐Ÿ‘

y = ax + b

8 = ๐Ÿ‘. 8 + b

b = -8 ๐Ÿ‘

y = ๐Ÿ‘ x โˆ’๐Ÿ– ๐Ÿ‘

y = ax + b

-2 = -๐Ÿ‘

๐Ÿ‘.6 + b

b = -2 + 2 ๐Ÿ‘y = -

๐Ÿ‘

๐Ÿ‘x -2 + 2 ๐Ÿ‘

Page 30: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 23.a) Como sรฃo paralelas, tem o mesmo coeficiente angular: ms = -2b) -2.ms = -1

ms = 1/2

c)โˆ’๐Ÿ“

๐Ÿ’. ms = -1

ms =๐Ÿ’

๐Ÿ“

d) Como sรฃo paralelas, tem o mesmo coeficiente angular: ms = 2/5

e)๐Ÿ

๐Ÿ•. ms = -1

ms =โˆ’๐Ÿ•

๐Ÿ

f)๐Ÿ’

๐Ÿ“. ms = -1

ms =โˆ’๐Ÿ“

๐Ÿ’

Page 31: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 24.

2y โ€“ x โ€“ 5 = 0

y = x/2 + 5/2

y = ax + b

2 = ๐Ÿ

๐Ÿ. 1 + b

b = ๐Ÿ‘

๐Ÿ

y = ๐Ÿ

๐Ÿx +

๐Ÿ‘

๐Ÿ

P(1, 2)

Page 32: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 25.

Intersecรงรฃo da reta S com o eixo das abcissas:x โ€“ y โ€“ 4 = 0x โ€“ 0 โ€“ 4 = 0x = 4 e y = 0

x + 2y + 2 = 02y = -x โ€“ 2y = (-1/2)x - 1

y = ax + b

0 = โˆ’๐Ÿ

๐Ÿ. 4 + b

b = 2

y = โˆ’๐ฑ

๐Ÿ+2

Page 33: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 26. C

Nas primeiras 2 horas, a variaรงรฃo foi de 1,5 KmNas 2 horas seguintes, a variaรงรฃo foi de 40 KmNas 2 horas finais, a variaรงรฃo foi de 10 KmO grรกfico que melhor representa estรก na letra C.

Page 34: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 27. C

60 + 40 + 60 + 40 + 20 + 80 = 300

Page 35: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 28.

A) ๐๐ซ๐žรง๐จ ๐ž๐ฆ๐ฆ๐š๐ซรง๐จ: ๐‘ท๐Ÿ‘

๐๐ซ๐žรง๐จ ๐ž๐ฆ ๐€๐›๐ซ๐ข๐ฅ: ๐‘ท๐Ÿ’

๐‘ท๐Ÿ’ = ๐‘ท๐Ÿ‘.(1 + 0,3)26 = ๐‘ท๐Ÿ‘.(1 + 0,3)๐‘ท๐Ÿ‘ = R$ 20,00

๐๐ซ๐žรง๐จ ๐ž๐ฆ๐ฆ๐š๐ซรง๐จ: ๐‘ท๐Ÿ‘

๐๐ซ๐žรง๐จ ๐ž๐ฆ ๐€๐›๐ซ๐ข๐ฅ: ๐‘ท๐Ÿ“

๐‘ท๐Ÿ“ = ๐‘ท๐Ÿ‘ . (1 + 0,56)๐‘ท๐Ÿ“= 20 . (1 + 0,56)๐‘ท๐Ÿ“ = R$ 31,20

B) ๐๐ซ๐žรง๐จ ๐ž๐ฆ๐ฆ๐š๐ข๐จ:๐‘$ ๐Ÿ‘๐Ÿ, ๐Ÿ๐ŸŽ

๐๐ซ๐žรง๐จ ๐ž๐ฆ ๐‰๐ฎ๐ง๐ก๐จ: ๐‘ท๐Ÿ” = ๐‘ท๐Ÿ‘ . (1 + 0,482) = R$ 29,64

ร๐ง๐๐ข๐œ๐ž: ๐ข

i = ๐Ÿ๐Ÿ—,๐Ÿ”๐Ÿ’ โˆ’๐Ÿ‘๐Ÿ,๐Ÿ๐ŸŽ

๐Ÿ‘๐Ÿ,๐Ÿ๐ŸŽ=

โˆ’๐Ÿ,๐Ÿ“๐Ÿ”

๐Ÿ‘๐Ÿ,๐Ÿ= โˆ’๐ŸŽ, ๐ŸŽ๐Ÿ“ = โˆ’๐Ÿ“% (๐‘น๐’†๐’…๐’–รงรฃ๐’)

Page 36: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 29. D

O nรบmero total de acidentes ocorridos รฉ 12 . 0 + 9 . 1 + 10 . 2 + 5 . 3 + 3 . 4 + 2 . 5 + 1 . 6 = 72. O nรบmero de motoristas que sofreram pelo menos quatro acidentes รฉ 3 + 2 + 1 = 6 > 5. O nรบmero de motoristas que sofreram no mรกximo dois acidentes รฉ 12 + 9 + 10 = 31 > 30.

Page 37: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 30. C

Pelo grรกfico temos que V = รกrea da figura formada no intervalo pedido,

entรฃo V = 1,4.(15 โ€“ 5) = 14 L.

Logo, Q = 4,8.14 = 67,2 kcal.

Page 38: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 31. D

O รบnico grรกfico que se passa retas verticais e nรฃo se toca em dois estรก na letra D. Logo รฉ uma funรงรฃo.

Page 39: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 32. E

f(-3/2) รฉ um valor entre 0 e 1.f(1/2) รฉ um valor entre 2 e 3.Essa soma sรณ poderรก estar entre 2 e 4.

Page 40: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 33.

A) De acordo com o grรกfico, entre 1940 e 1950

B) ๐Ÿ๐ŸŽ โˆ’๐Ÿ•

๐Ÿ–๐ŸŽ โˆ’๐Ÿ”๐ŸŽ= ๐ŸŽ,๐Ÿ๐Ÿ“ ๐’Ž๐’Š๐’ ๐’‰๐’‚๐’ƒ๐’Š๐’•๐’‚๐’๐’•๐’†๐’” ๐’‘๐’๐’“ ๐’‚๐’๐’

Sabendo-se que na dรฉcada de 80 a populaรงรฃo รฉ de 10 mil, para se chegar a 20 mil faltam 10 mil pessoas.๐Ÿ๐ŸŽ

๐ŸŽ, ๐Ÿ๐Ÿ“โ‰ˆ ๐Ÿ”๐Ÿ• ๐’‚๐’๐’๐’”

Logo, 1980 + 67 = 2047Resposta: Entre 2040 e 2050

Page 41: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 34. B

B) Falsa, pois depois de uma certa quantidadeingerida a absorรงรฃo se mantรฉm constante.

Page 42: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 35. D

Page 43: Resoluรงรฃo Meta 6 - Darlan Moutinhoa) (r): 2x โ€“3y + 7 = 0 e P(2, 3) 2x โ€“3y + 7 = 0 y = 2๐‘ฅ 3 +7 3 Como as retas sรฃo perpendiculares: 2 3. =โˆ’ = โˆ’3 2 y = ax + b 3 = โˆ’3

QUESTรƒO 36. E

e)f(2) + f(3) = 2 + 3 = 5f(5) = 4Logo, f(2) + f(3) โ‰  f(5)


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