All categories

3 years ago

Factor the following equation to find its zeros. y = x² - 36

Mathematics

1 answer:

alukav5142 [94]3 years ago

3 0

Answer:

x = 6, x = - 6

Step-by-step explanation:

Given

y = x² - 36

To find the zeros let y = 0, that is

x² - 36 = 0 ← x² - 36 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b), thus

x² - 36 = 0

x² - 6² = 0

(x - 6)(x + 6) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 6 = 0 ⇒ x = - 6

You might be interested in

EB bisects ∠RED. Find the measure of ∠BED if ∠REB = 5x – 3 and ∠RED = 8x + 16<br><br>plzzzzzzzzzzz

Allushta [10]

Answer:

this is 52 degrees

Step-by-step explanation:

7 0

2 years ago

Pls help ill give brainliest :)

Tatiana [17]

Answer:

380.

Just add the two numbers together and it'll work :)

3 0

2 years ago

Also explain how you got your answer please

serg [7]

Answer:

Step-by-step explanation:

8 0

3 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:

$P(X > x) = 0.08$