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steposvetlana [31]
3 years ago
15

Identify the zeros of the polynomial function N(x)=1/2 (x-1)(x+3).

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
6 0

Answer:

a:x=-3

c:x=1

Step-by-step explanation:

The zeros of a function are the values of x for which the value of the function f(x) becomes zero.

In this problem, we have the following function:

f(x)=\frac{1}{2}(x-1)(x+3)

Here we want to find the zeros of the function, i.e. the values of x for which

f(x)=0

In order to make f(x) equal to zero, either one of the factors (x-1) or (x+3) must be equal to zero.

Therefore, the two zeros can be found by requiring that:

1)

x-1=0\\\rightarrow x=+1

2)

x+3=0\\\rightarrow x=-3

So the correct options are

a:x=-3

c:x=1

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2.) Mila completed a half marathon race over the weekend. The full race was
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2 years ago
A store has been selling 100 Blu-ray disc players a week at $600 each. A market survey indicates that for each $40 rebate offere
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Answer:

<em>The rebate should be $220</em>

Step-by-step explanation:

<u>Demand Curve</u>

It's the relationship between price (P) and quantity (Q) demanded a certain product or service.

(a) We need to find the function that relates both magnitudes assuming a linear equation. The equation of a line can be found with the point-point formula:

\displaystyle Q-Q_1=\frac{Q_2-Q_1}{P_2-P_1}(P-P_1)

Two sets of data are given: 100 Blu-ray disc players are sold a week at $600 each. The ordered pair for this condition is (P,Q)=(600,100).

The other point comes from the market survey: The number of units sold will increase by 80 (100+80=180) when the price goes down $40 (600-40=560). The new point is (P,Q)=(560,180)

We set up the equation of the demand

\displaystyle Q-100=\frac{180-100}{560-600}(P-560)

Rearranging

-40Q+4000=80P-44800

Or

80P+40Q=48800

Simplifying

2P+Q=1120

(b) The revenue function is Q times the price

R=Q.P

Solving the equation of the demand for P

\displaystyle P=\frac{1120-Q}{2}

Thus, the revenue is

\displaystyle R=Q\cdot \frac{1120-Q}{2}

\displaystyle R=\frac{1120Q-Q^2}{2}

(c) To find the optimum value of the revenue, we take the derivative of R and equate to 0

\displaystyle R'=\frac{1120-2Q}{2}=560-Q=0

Solving

Q=560 units a week

For which the revenue is

\displaystyle R=\frac{1120(560)-560^2}{2}

R=\$156,800

And the price is

\displaystyle P=\frac{1120-560}{2}=280

P=\$280

The rebate should be $600-$280=$220

4 0
3 years ago
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