All categories

2 years ago

How is a marketing -oriented firm different from a production-oriented firm or a sales-oriented firm?

Mathematics

1 answer:

dedylja [7]2 years ago

4 0

Marketing-oriented firms focus more on what the market needs above every other aspects.

<h3>What is a Marketing-oriented Firm?</h3>

Marketing-oriented firms can be described as firms that major in identifying the needs and wants of customers and then creating specific products that are designed to meet the wants of such customers. One of the important elements of marketing-oriented firms is to ensure there is a demand for whatever products or services they offer.

Some examples of marketing-oriented firms are:

Apple
Coca-Cola
Amazon

<h3>What is a Product-oriented Firm or a Sales-oriented Firm?</h3>

A product-oriented firm focus more resources on and focus on researching and developing quality products rather than focusing more on the market to discover the wants of customers.

Sales-oriented firm tend to focus more on developing its sales force that will promote their products and services.

<h3>How Marketing-Oriented Firms are Different from the Product-oriented or Sales-oriented Firms?</h3>

Marketing-oriented firms stand out and are different from the other two types of firms because they focus more on what the market needs above every other aspects.

Learn more about Marketing-oriented firms on:

brainly.com/question/13288643

#SPJ1

You might be interested in

If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$ . Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$