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masya89
2 months ago
10

The steps in writing f(x) = 18x + 3x2 in vertex form are shown, but a value is missing in the last step. Write the function in s

tandard form. f(x) = 3x2 + 18x Factor a out of the first two terms. f(x) = 3(x2 + 6x) Form a perfect square trinomial. f(x) = 3(x2 + 6x + 9) – 3(9) Write the trinomial as a binomial squared. f(x) = 3(x + ______)2 – 27
Mathematics
2 answers:
lawyer [12.5K]2 months ago
7 0

To determine the vertex form of the given parabola y=a(x-x_0)^2+y_0 nearly in standard form y=ax^2+bx+c,the following steps need to be observed:

1. Convert the function into standard form:

y=3x^2+18x.

2. Factor out the coefficient from the first two terms:

y=3(x^2+6x).

3. Create a perfect square trinomial:

y=3(x^2+6x+9-9)=3(x^2+6x+9)-3\cdot 9.

4. Present the trinomial as a binomial squared:

y=3(x+3)^2-27.

The vertex coordinates are (-3,-27).

Result: the value that is missing is 3

PIT_PIT [12.4K]2 months ago
5 0

Response:

It appears to be A on e2020

Explanation step-by-step:

You might be interested in
What is the empirical formula of an oxide of chromium that is 48 percent oxygen by mass?
Svet_ta [12734]

Answer:

CrO₃.

Step-by-step explanation:

First, we will find the chromium percentage in the oxide by using the following process:

Oxygen (O) = 48%

Chromium (Cr) =?

The oxide consists solely of chromium and oxygen, and its chromium percentage is calculated as:

Cr = 100 – percentage of oxygen

Cr = 100 – 48

Cr = 52%

Next, we will determine the empirical formula for the oxide:

Chromium (Cr) = 52%

Oxygen (O) = 48%

Now, we divide by their molar mass:

Cr = 52/52 = 1

O = 48/16 = 3

Now, divide by the lowest value:

Cr = 1/1 = 1

O = 3/1 = 3

Thus, the empirical formula for the oxide is CrO₃.

4 0
2 months ago
Two car rental companies charge an initial rental fee plus a fee per day. The table shows the total costs of renting a car for x
AnnZ [12381]

Answer:

a) Which rental company charges a lower daily rate?

Payless rental company.

b) How many days must you rent a car for the total costs to equal?​

3 days.

Step-by-step breakdown:

We are comparing two rental services:

Discount Car Rentals and Payless Car Rentals.

x = rental duration in days

y = total cost.

a) Assuming the rental period starts at 2 days:

We need to formulate the equation for Discount Rentals.

For Discount Car Rentals:

2 days = $185

4 days = $265

6 days = $345

8 days = $425

The cost differential for 2 days = $265 - $185 = $80

Therefore, for 1 day the cost difference is = $80/2

= $40

For a one-day rental,

185 - 40 = $145

Thus, the equation for the payment of Discount Rentals is:

y = 40x + 105

b) For Payless Car Rentals:

The provided equation is: y = 55x + 60.

y = total cost, x = rental days

a) For 1 day:

y = 55(1) + 60

y = $115

b) For 2 days:

y = 55(2) + 60

y = 110 + 60

y = $170

c) For 3 days:

y = 55(3) + 60

= 165 + 60

= $225

d) For 4 days:

y = 55x + 60.

y = 55 × 4 + 60

y = 220 + 60

y = $280

e) For 5 days:

y = 55x + 60.

y = 55 × 5 + 60

y = 275 + 60

y = $335

f) For 6 days:

y = 55x + 60.

y = 55 × 6 + 60

y = 330 + 60

y = $380

g) For 7 days:

y = 55x + 60.

y = 55 × 7 + 60

y = 385 + 60

y = $445

h) For 8 days:

y = 55x + 60.

y = 55 × 8 + 60

y = 440 + 60

y = $500

a) Which rental company charges a lower daily rate?

The total expense of renting a vehicle as calculated above:

For 1 day from Discount Car Rentals = $145

For 1 day from Payless Car Rentals = $115

Based on these calculations, Payless rental company offers the lower rate.

b) For how many days is the rental required to have costs equal?

We set both equations equal to each other:

y = 40x + 105 = y = 55x + 60.

40x + 105 = 55x + 60

105 - 60 = 55x - 40x

45 = 15x

x = 45/15

x = 3 days.

Hence, after 3 days, the total cost of renting a vehicle from either Discount or Payless Car Rentals will be equal.

8 0
2 months ago
Graph the image of the given triangle after the transformation with the rule (x, y)→(x, −y) .
Zina [12379]
To begin, we identify three corner points: A=(2,3), B=(6,8), and C=(7,4). We apply the transformation rule (x, y)→(x, −y), reflecting across the x-axis. This means we will invert the y-values. The new coordinates are: A=(2,-3), B=(6,-8), and C=(7,-4). Subsequently, we can plot these points and create a graph.
4 0
1 month ago
Read 2 more answers
In a study of exercise, a large group of male runners walk on a treadmill for six minutes. Their heart rates in beats per minute
Leona [12618]

Answer:

a) 1.88% de los corredores tienen frecuencias cardíacas superiores a 130.

b) 50% de los no corredores tienen frecuencias cardíacas superiores a 130.

Explicación paso a paso:

Los problemas relacionados con muestras distribuidas normalmente pueden resolverse utilizando la fórmula del puntaje z.

En un conjunto con media \mu y desviación estándar \sigma, el puntaje z asociado a una medida X se presenta a través de:

Z = \frac{X - \mu}{\sigma}

El puntaje z indica cuántas desviaciones estándar está la medida respecto a la media. Tras determinar el puntaje z, consultamos la tabla de puntajes z y localizamos el valor p relacionado con este puntaje. Este valor p representa la probabilidad de que el valor de la medida sea menor que X, es decir, el percentil de X. Al restar 1 del valor p, obtenemos la probabilidad de que el valor de la medida sea mayor que X.

(a) ¿Qué porcentaje de los corredores tienen frecuencias cardíacas superiores a 130?

En un estudio sobre ejercicio, un gran grupo de corredores masculinos camina en una caminadora durante seis minutos. Sus frecuencias cardíacas expresadas en latidos por minuto al final varían entre ellos conforme a la distribución N(104,12.5). Esto implica que \mu = 104, \sigma = 12.5.

Este porcentaje es 1 menos el valor p cuando X = 130.

Z = \frac{X - \mu}{\sigma}

Z = \frac{130 - 104}{12.5}

Z = 2.08

Z = 2.08 tiene un valor p de 0.9812.

Esto significa que 1-0.9812 = 0.0188 = 1.88% de los corredores tienen frecuencias cardíacas superiores a 130.

(b) ¿Qué porcentaje de los no corredores tienen frecuencias cardíacas superiores a 130?

Las frecuencias cardíacas para hombres no corredores después del mismo ejercicio mantienen la distribución N(130, 17). Esto implica que \mu = 130, \sigma = 17.

Este porcentaje es 1 menos el valor p cuando X = 130.

Z = \frac{X - \mu}{\sigma}

Z = \frac{130 - 130}{17}

Z = 0.00

Z = 0.00 tiene un valor p de 0.5000.

Esto significa que 1-0.50 = 0.50 = 50% de los no corredores tienen frecuencias cardíacas superiores a 130.

4 0
2 months ago
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