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3 years ago

The graph of function g is a vertical stretch of the graph of function f by a factor of 3.

Mathematics

1 answer:

Assoli18 [71]3 years ago

7 0

Question says that the graph of function g is a vertical stretch of the graph of function f by a factor of 3.

Now we have to find about which equation describes function g.

We know that if f(x) is the given function then a*f(x) gives vertical stretch or compress by a factor of a.

when 0<a<1 then it compresses by factor of a.

when a>1 then it stretches vertically by factor of a.

Given that function f is stretched by a factor of a so that means we use a=3 in formula a*f(x)

Hence final answer will be g(x)=3f(x).

So choice A is correct answer.

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3 years ago

The average number of customers per day increased 25 percent during the sale. If the average number of customers per day before

Rasek [7]

Answer:

The average number of customers per day during the sale was 150.

Step-by-step explanation:

This question can be solved using a rule of three.

Before the sale, the average number of customers per day was 120, which is 100% = 1.

Now, there are x customers, and since it is an increase of 25%, it is 100+25 = 125% = 1.25. So

120 - 1

x - 1.25

x = 1.25*120 = 150

The average number of customers per day during the sale was 150.

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3 years ago

The product of four consecutive odd numbers(all four positive or all four negative)is 62985.find the two possible sets of four n

emmasim [6.3K]

Answer:

Step-by-step explanation:

Let X be the first number.

(X)*(X+2)*(X+4)*(X+6) = 62985

(X^2 + 6X)^2 + 8(X^2 + 6X) = 62985

X = 13, AND X= -19

A. X = 13

(13)(15(17)(19) = 62985

B. X = -19

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2 years ago

What is a cow well 11111111111111111111111111111111111111

dybincka [34]

Answer: it is all of them

Step-by-step explanation:

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3 years ago

Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?

Naddik [55]

El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.

<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>

La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.

Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:

σ = W / (π · D² / 4)

Donde:

W - Peso de la caja, en newtons.
D - Diámetro del área transversal de la caja, en metros.

Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:

σ = 25 N / [π · (0.02 m)² / 4]

σ ≈ 79577.472 Pa

<h3>Observación</h3>

La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.

Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145

#SPJ1

5 0

1 year ago

The graph of function g is a vertical stretch of the graph of function f ​​by a factor of 3.

The graph of function g is a vertical stretch of the graph of function f by a factor of 3.