The y2kcenter.org is 14.1345/99 ≈ 13.58Let”s round it to the one decimal:13.58 ≈ 13.6Let”s round it to the nearest integer:13.6 ≈ 14

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The set of ordered pairs (-1, 8), (0, 3), (1.-2), and (2, -7) represent a function. What is the range of the function?

First, the range is all about the y-values, so the set notation needs to have the form:{ y : y __________} Because of that, you can eliminate the first and third options that use “x”. Using x would reference the domain.Second, the range is the set of all y-values used in the function. You only include y-values in the range, so you’d only include 8, 3, -2, and -7 in your y2kcenter.org.That means it’s the second y2kcenter.org.Side note:Some areas would use the notation { y | y = -7, -2, 3, 8}, using “|” instead of “:” in the set notation.f

5 0

9 months ago

The two sales people for a local advertising firm are Trinity and Jason. Trinity sold $2,330 in ads and Jason sold $1,740. What

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pav-90 <236>

The two sales people for a local advertising firm are Trinity and Jason.

Sales amount of Trinity = $2330

Sales amount of Jason = $1740

Total sales of both the people =

The fractional value of Trinity”s sales =

or 57%

The fractional value of Jason”s sales =

or 43%

5 0

11 months ago

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Nostrana <21>

**y2kcenter.org:**

1. x = 02. x = 0 and x = 3/43. x = – 2 and x = – 34. n = 4 and n = 15. x = -1 and x = 2/36. x = -1 and x = 1/27. Problem 4: x = 10 cm

**Explanation:**

1. 4x²=0

a) Divide both sides by 4:

x² = 0

b) Take square root of both sides

x = 0

c) Solution: there is only one solution: x = 0.

d) Substitute to verify:

4(0)² = 04(0) = 00 = 0

2. 4x² – 3x = 0

a) Common factor x:

x(4x – 3) = 0

b) Zero product property: any or both factors must equal zero

i) x = 0

ii) 4x – 3 = 0

4x = 3

x = 3/4

Solutions x = 0 and x = 3

c) Verify the solutions:

i) x = 0 ⇒ 4(0)² – 3(0) = 0 – 0 = 0

ii) x = 3/4 ⇒ 4(3/4)² – 3(3/4) =3²/4 – 9/4 = 9/4 – 9/4 = 0

3. x² + 5x + 6 = 0

a) Factor: find two numbers whose sum is 5 and product is 6: 3 and 2:

(x + 2)(x + 3) = 0

b) Zero product property:

i) x + 2 = 0

x = – 2

ii) x + 3 = 0

x = – 3

The two solutions are x = – 2 and x = -3

c) Verify

i) x = – 2 ⇒ (-2)² + 5(-2) + 6 = 4 – 10 + 6 = – 6 + 6 = 0

ii) x = – 3 ⇒ (-3)² + 5(-3) + 6 = 9 – 15 + 6 = 9 – 9 = 0

4. n² – 5n + 4 = 0

a) Factor: find two numbers whose sum is -5 and product is 4: -4 and -1

(n – 4)(n – 1) = 0

b) Zero product property:

i) n – 4 = 0 ⇒ n = 4

ii) n – 1 = 0 ⇒ n = 1

The two solutions are n = 4 and n = 1

c) Verify the solutions:

i) n = 4 ⇒(4)² – 5(4) + 4 = 16 – 20 + 4 = 16 – 16 = 0

ii) n = 1 ⇒ (1)² – 5(1) + 4 = 1 – 5 + 4 = 1 – 1 = 0

5. 3x² + x – 2 = 0

a) Substitute x with 3x – 2x

3x² + 3x – 2x – 2 = 0

b) Group the terms (associative property)

(3x² + 3x) – (2x + 2)

c) Factor each group:

3x(x + 1) – 2(x + 1) = 0

d) Common factor x + 1:

(x + 1) (3x – 2) = 0

e) Zero product property

i) x + 1 = 0 ⇒ x = – 1

ii) 3x – 2 = 0 ⇒ x = 2/3

The two solutions are x = -1 and x = 2/3

f) Verify the solutions:

i) x = – 1 ⇒ 3(-1)² + (-1) – 2 = 3 – 1 – 2 = 2 – 2 = 0

ii) x = 2/3 ⇒ 3(2/3)² + (2/3) – 2 = 4/3 + 2/3 – 2 = 6/3 – 2 = 0

6. 6x² + 3x – 3 = 0

a) Divide both side by 3:

2x² + x – 1 = 0

b) Substiture x with 2x – x

2x² + 2x – x – 1 = 0

c) Group terms:

(2x² + 2x) – (x + 1) = 0

d) Factor each binomial

2x(x + 1) – (x + 1) = 0

e) Common factor x + 1

(x + 1)(2x – 1) = 0

f) Zero product property

i) x + 1 = 0 ⇒ x = – 1

ii) 2x + 1 = 0

x = 1/2

The two solutions are x = – 1 and x = 1/2

e) Verify

i) x = – 1 ⇒ 6(-1)² + 3(-1) – 3 = 6 – 3 – 3 = 0

ii) x = 1/2 ⇒ 6(1/2)² + 3(1/2) – 3 = 6/4 + 3/2 – 3 = 3 – 3 = 0

7. Problem

*4. Dado el siguiente rectángulo, determina cuál debe ser el valor de x que permita obtener el área que se indica. *

The area of a rectangle is the product of the length and the width:

Area = (x – 4)(x + 1) = 66

Solve the equation:

a) Distributive property:

x² + x – 4x – 4 = 66

b) Add like terms:

x² -3x – 4 = 66

c) Subtract 66 from both sides

x² -3x – 70 = 0

d) Factor: find two numbers whose sum is -3 and product is – 70: – 10 and 7

(x – 10)(x + 7) = 0

e) Zero product property:

i) x – 10 = 0 ⇒ x = 10

ii) x + 7 = 0 ⇒ x = -7

x = – 7 is not a valid solution because that would make the side lengths negative.