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3 years ago

Solve the equation for all values of x by completing the square. 3x^2-60x+255=0

Mathematics

1 answer:

erik [133]3 years ago

4 0

Answer:

x=5 and x=15

Step-by-step explanation:

3x^2-60x+225=0

3x^2-60x+225/3=0/3

x^2-20x+75=0

x^2-5x-15x+75=0

x(x-5)-15(x-5)=0

(x-5)(x-15)=0

x-5=0 x-15=0

x=5 x=15

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If a,b,c,d, and e are non negative such that x^6-289x^4-x^2+289=(x-a)(x+b)(x_c)(x+d)(x^2+ex+1), what is the value of a+b+c+d+e?

Irina18 [472]

Answer:

Step-by-step explanation:

Hello,

If $X=x^2$ it becomes

$X^3-289X^2-X+289=(X^2-1)(X-289)$

and $289=17^2$

$X^2-1=(X-1)(X+1)=(x^2-1)(x^2+1)=(x+1)(x-1)(x^2+1) \\\\x^2-289=x^2-17^2=(x-17)(x+17)$

So,

$x^6-289x^4-x^2+289=(x^2-1)(x^2+1)(x^2-289)=(x-1)(x+1)(x+17)(x-17)(x^2+1)$

a = 1

b = 1

c = 17

d = 17

e = 0

Then a + b + c + d + e = 36

Thanks

3 0

3 years ago

An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba

madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688$

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

$P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312$

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0

3 years ago

Convert the fraction greater than one into a mixed number 9/4

Pani-rosa [81]

Answer: 2 1/4

Step-by-step explanation:

7 0

3 years ago

Diego runs for 32 seconds at -8.1 meters per second. What is his finish point?

nalin [4]

Answer:

32×(-8.1)=-259.2

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A number increased by 5 is less than or equal to 26

blagie [28]

Answer:

Equation: x + 5 ≤ 26

Solved: x ≤ 21

3 0

3 years ago