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2 years ago

The function y = - 4x + 14 gives the number y of avocados you have left after making x batches of

Mathematics

1 answer:

GuDViN [60]2 years ago

7 0

Answer:

See below

Step-by-step explanation:

A. x is independent and y is dependent

B The range is 14, 10, 6, and 2

D The domain is discrete because it only includes whole avocados, and not fractional pieces.

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3 years ago

When 10 b = 5 (StartRoot c EndRoot + 2) is solved for c, one equation is c = (2b minus 2) squared Which of the following is an e

Burka [1]

Answer:

I will show all the steps to lead to the solution.

1) 10b = 5 (√c + 2) <---- starting equation

2) divide by 5 => 2 b = √c + 2

3) subtract 2 => 2b - 2 = √c

4) raise to the power 2 => (2b - 2)^2 = c, which is the expression given.

Now these are two equivalent forms:

1) extract 2 from the parentheses: c = 4 (b -1)^2

2) expand the parentheses: c = 4b^2 - 8b + 4)

Step-by-step explanation:

Choice D

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3 years ago

If X²⁰¹³ + 1/X²⁰¹³ = 2, then find the value of X²⁰²² + 1/X²⁰²² = ?

enyata [817]

Step-by-step explanation:

$\bf➤ \underline{Given-} \\$

$\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\$

$\bf➤ \underline{To\: find-} \\$

$\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\$

$\bf ➤\underline{Solution-} \\$

Let us assume that:

$\rm: \longmapsto u = {x}^{2013}$

Therefore, the equation becomes:

$\rm: \longmapsto u + \dfrac{1}{u} = 2$

$\rm: \longmapsto \dfrac{ {u}^{2} + 1}{u} = 2$

$\rm: \longmapsto{u}^{2} + 1 = 2u$

$\rm: \longmapsto{u}^{2} - 2u + 1 =0$

$\rm: \longmapsto {(u - 1)}^{2} =0$

$\rm: \longmapsto u = 1$

Now substitute the value of u. We get:

$\rm: \longmapsto {x}^{2013} = 1$

$\rm: \longmapsto x = 1$

Therefore:

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 1 + 1$

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 2$

★ Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3ab(a - b) - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x - b) = x² + (a - b)x - ab

(x - a)(x + b) = x² - (a - b)x - ab

(x - a)(x - b) = x² - (a + b)x + ab

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3 years ago