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The collection of ordered pairs (-1, 8), (0, 3), (1.-2), and also (2, -7) stand for a function. What is the range of the function?

First, the variety is all around the y-values, therefore the collection notation needs to have actually the form: y : y __________ due to the fact that of that, you can get rid of the first and third options that use “x”. Utilizing x would reference the domain.Second, the selection is the collection of all y-values offered in the function. You only encompass y-values in the range, so you’d only incorporate 8, 3, -2, and also -7 in her aramuseum.org.That way it’s the 2nd aramuseum.org.Side note:Some locations would use the notation y , utilizing “|” instead of “:” in the set notation.f

5 0

8 months ago

The two sales human being for a local heralding firm room Trinity and Jason. Trinity offered $2,330 in ads and also Jason offered $1,740. What

pav-90 <236>

The two sales world for a local advertising firm room Trinity and Jason.

Sales quantity of Trinity = $2330

Sales quantity of Jason = $1740

Total sales that both the world =

The fractional value of Trinity"s sales =

or 57%The fractional worth of Jason"s sales =

or 43%5 0

10 month ago

Read 2 an ext aramuseum.orgs

ES URGENTE!!! POR donate AYUDA!!!

See more: On A Topographic Map, What Does It Mean When Contour Lines Are Close Together

Nostrana <21>

**aramuseum.org:**

**Explanation:**

1. 4x²=0

a) divide both sides by 4:

x² = 0b) take square source of both sides

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

d) substitute to verify:

4(0)² = 04(0) = 00 = 02. 4x² - 3x = 0

a) common factor x:

x(4x - 3) = 0b) Zero product property: any or both determinants 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 amount is 5 and also product is 6: 3 and 2:

(x + 2)(x + 3) = 0b) Zero product property:

i) x + 2 = 0

x = - 2

ii) x + 3 = 0

x = - 3

The two remedies 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: uncover two number whose amount is -5 and also product is 4: -4 and also -1

(n - 4)(n - 1) = 0b) Zero product property:

i) n - 4 = 0 ⇒ n = 4

ii) n - 1 = 0 ⇒ n = 1

The two remedies 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 through 3x - 2x

3x² + 3x - 2x - 2 = 0b) group the terms (associative property)

(3x² + 3x) - (2x + 2)c) variable each group:

3x(x + 1) - 2(x + 1) = 0d) typical factor x + 1:

(x + 1) (3x - 2) = 0e) 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 = 0b) Substiture x with 2x - x

2x² + 2x - x - 1 = 0c) team terms:

(2x² + 2x) - (x + 1) = 0d) element each binomial

2x(x + 1) - (x + 1) = 0e) usual factor x + 1

(x + 1)(2x - 1) = 0f) 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 the a rectangle is the product of the length and the width:

Area = (x - 4)(x + 1) = 66Solve the equation:

a) Distributive property:

x² + x - 4x - 4 = 66b) add like terms:

x² -3x - 4 = 66c) Subtract 66 native both sides

x² -3x - 70 = 0d) Factor: discover two numbers whose sum is -3 and also product is - 70: - 10 and also 7

(x - 10)(x + 7) = 0e) Zero product property:

i) x - 10 = 0 ⇒ x = 10

ii) x + 7 = 0 ⇒ x = -7

x = - 7 is not a precious solution since that would make the next lengths negative.