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3 years ago

Given the functions j(x) = x2 − 9 and k(x) = −x + 7, which operation results in a 3rd degree polynomial?

Mathematics

1 answer:

sveta [45]3 years ago

7 0

So when you multiply j(x) with k(x), the first term will be -x^3. So multiplication.

I hope this helps

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Look at the function f(x) = x3 - 4 x. which describes an interval in which the function is increasing?

Bad White [126]

Differentiate it
get f ' (x)=3x^2-4
put 3x^2-4=0
x=2/√3,-2/√3
so interval
x∈(-2/√3,2/√3)

5 0

3 years ago

This famous clock tower has a clock face with a diameter of 23 feet. What is the approximate area of this clock face

pashok25 [27]

Answer:

Therefore the approximate area of this clock face is 415.3 feet².

Step-by-step explanation:

Given:

Clock is of Circular Shape,

Diameter = 23 feet

$Radius =r =\dfrac{diameter}{2}=\dfrac{23}{2}=11.5\ feet$

To Find :

Area of Clock Face = ?

Solution:

Area of Circle having Radius 'r' is given as,

$\textrm{Area of Circle}=\pi r^{2}$

Substituting the values we get

$\textrm{Area of Clock Face}=3.14\times (11.5)^{2}=415.265=415.3\ feet^{2}$

Therefore the approximate area of this clock face is 415.3 feet².

7 0

3 years ago

Can someone plz help me with this one problem plzzzzz!!! I will mark brainliest

Tpy6a [65]

Answer:

2 = 76

3 = 114

8 = 304

10 = 380

5 0

2 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$