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11Alexandr11 [23.1K]
2 years ago
8

A company makes greeting cards and their research shows that that price and demand are related linearly: p=mx +b.They know that

for every additional card they wish to sell they need to drop the price by $0.05. They also know that in order to sell 300 cards they need to set the price at $7. Find the linear equation relating P price to demand. Preview p =
Mathematics
1 answer:
BARSIC [14]2 years ago
8 0

Answer:

P= -0.05q+22

Step-by-step explanation:

To find the linear equation that relates price with quantity demanded, first we must find the slope. Because the independent variable is the quantity demanded and the dependent variable is the price, the slope represents how the price changes when there is an extra unit of quantity demanded. The problem gives this information: "for every additional card (extra unit) they need to drop the price by $0.05". The slope (m) in this case is negative because an extra unit, reduces the price: -0.05

The second step is to use this formula:

Y-y1= m*(X-x1)

y1 and x1 is a point of the demand curve, in this case it is y1= $7 and x1=300

Y-$7= -$0.05*(X-300)

Y-7=-0.05X+15

Y= -0.05X+15+7

Y= -0.05X-22

Price= -0.05 quantity demanded +22

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F(x)=2x-3 g(x)=x^2+1 find g(2)
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Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function f(
alekssr [168]

Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function ()=−(+)^−

<em><u>Answer:</u></em>

vertex = (-4, -5)

Axis of symmetry = -4

use the (-4, -5) to find the minimum value

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

<em><u>Solution:</u></em>

Given function is:

f(x) = (x+4)^2 - 5

The equation in vertex form is given as:

y = a(x-h)^2+k

Where, (h, k) is constant

On comparing give function with vertex form,

h = -4

k = -5

Vertex is (-4 , -5)

Axis of symmetry : x co-ordinate of vertex

Thus, axis of symmetry = -4

The coefficient of x^2 is positive in given function.

Thus the vertex point will be a minimum

Minimum\ value = f(\frac{-b}{a})

f(x) = x^2 + 8x + 16 - 5\\\\f(x) = x^2 + 8x + 11

f(x) = ax^2+bx+c

On comparing,

a = 1

b = 8

x = \frac{-b}{2a} = \frac{-8}{2 \times 1} = -4

f(-4) = (-4)^2 + 8(-4) + 11 = 16 - 32 + 11 = -5

Thus, use the (-4, -5) to find the minimum value

Domain and range

f(x) = (x+4)^2 - 5

The domain is the input values shown on the x-axis

The range is the set of possible output values f(x)

Therefore,

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

4 0
2 years ago
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