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Advocard [28]
3 years ago
9

avocados cost $3 per pound. write and graph an equation in two variables that represents the cost of buying avocados

Mathematics
1 answer:
EastWind [94]3 years ago
4 0
Pounds will be represented by x and the cost will be represented by y. The equation is Y=3x. For the graph just use substition for x and then calculate the outcome and graph it.
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(08.06 MC)
const2013 [10]

Answer:

  (3.5, 17)

Step-by-step explanation:

It would be nice to see the whole graph, so we can see where the functions cross.

Without that information, we can still eliminate unreasonable choices.

A) the quadratic at y=3.5 is well above the exponential

B) the most likely choice (3.5, 17)

C) at x=-8, the quadratic is above the exponential

D) neither graph goes anywhere near y = -8

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La potencia que se obtiene de elevar a un mismo exponente un numero racional y su opuesto es la misma verdadero o falso?
malfutka [58]

Answer:

Falso.

Step-by-step explanation:

Sea d = \frac{a}{b} un número racional, donde a, b \in \mathbb{R} y b \neq 0, su opuesto es un número real c = -\left(\frac{a}{b} \right). En el caso de elevarse a un exponente dado, hay que comprobar cinco casos:

(a) <em>El exponente es cero.</em>

(b) <em>El exponente es un negativo impar.</em>

(c) <em>El exponente es un negativo par.</em>

(d) <em>El exponente es un positivo impar.</em>

(e) <em>El exponente es un positivo par.</em>

(a) El exponente es cero:

Toda potencia elevada a la cero es igual a uno. En consecuencia, c = d = 1. La proposición es verdadera.

(b) El exponente es un negativo impar:

Considérese las siguientes expresiones:

d' = d^{-n} y c' = c^{-n}

Al aplicar las definiciones anteriores y las operaciones del Álgebra de los números reales tenemos el siguiente desarrollo:

d' = \left(\frac{a}{b} \right)^{-n} y c' = \left[-\left(\frac{a}{b} \right)\right]^{-n}

d' = \left(\frac{a}{b} \right)^{(-1)\cdot n} y c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{(-1)\cdot n}

d' = \left[\left(\frac{a}{b} \right)^{-1}\right]^{n}y c' = \left[(-1)^{-1}\cdot \left(\frac{a}{b} \right)^{-1}\right]^{n}

d' = \left(\frac{b}{a} \right)^{n} y c = (-1)^{n}\cdot \left(\frac{b}{a} \right)^{n}

d' = \left(\frac{b}{a} \right)^{n} y c' = \left[(-1)\cdot \left(\frac{b}{a} \right)\right]^{n}

d' = \left(\frac{b}{a} \right)^{n} y c' = \left[-\left(\frac{b}{a} \right)\right]^{n}

Si n es impar, entonces:

d' = \left(\frac{b}{a} \right)^{n} y c' = - \left(\frac{b}{a} \right)^{n}

Puesto que d' \neq c', la proposición es falsa.

(c) El exponente es un negativo par.

Si n es par, entonces:

d' = \left(\frac{b}{a} \right)^{n} y c' = \left(\frac{b}{a} \right)^{n}

Puesto que d' = c', la proposición es verdadera.

(d) El exponente es un positivo impar.

Considérese las siguientes expresiones:

d' = d^{n} y c' = c^{n}

d' = \left(\frac{a}{b}\right)^{n} y c' = \left[-\left(\frac{a}{b} \right)\right]^{n}

d' = \left(\frac{a}{b} \right)^{n} y c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{n}

d' = \left(\frac{a}{b} \right)^{n} y c' = (-1)^{n}\cdot \left(\frac{a}{b} \right)^{n}

Si n es impar, entonces:

d' = \left(\frac{a}{b} \right)^{n} y c' = - \left(\frac{a}{b} \right)^{n}

(e) El exponente es un positivo par.

Considérese las siguientes expresiones:

d' = \left(\frac{a}{b} \right)^{n} y c' = \left(\frac{a}{b} \right)^{n}

Si n es par, entonces d' = c' y la proposición es verdadera.

Por tanto, se concluye que es falso que toda potencia que se obtiene de elevar a un mismo exponente un número racional y su opuesto es la misma.

3 0
3 years ago
Given values of a data set:
castortr0y [4]

Answer:

d

Step-by-step explanation:

7 0
3 years ago
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