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3 years ago

Using the polynomials Q = 3 x2 + 5 x - 2, R = 2 - x2 , and S = 2 x + 5, perform the indicated operation.

Mathematics

1 answer:

marusya05 [52]3 years ago

4 0

Answer:

4x^2 + 3x - 9

Step-by-step explanation:

First, add together functions R(x) and S(x). Stack them vertically and combine like terms:

R(x) = 2 - x^2

S(x) = 2x + 5

------------------

(R+S)(x) = -x^2 + 2x + 7

Now subtract the above result from Q(x):

Q(x) - (R(x) + S(x)) = 3x^2 + 5x - 2 + x^2 - 2x - 7, or 4x^2 + 3x - 9

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The sum of a number y and 9 is a least -1

Nataly_w [17]

Answer:

Y+9>= -1

Step-by-step explanation:

The sum of a number y and 9 means that we wanna add Y and 9, but since y is a variable and 9 is just a number, we can't add them.

Now we have Y+9

Now, we plug in our inequality sign which is at least.

at least is >=

Now we have Y+9>=

And all you have to do is place the negative one after the inequality sign, and you get Y+9>= -1

Hope this helps! :)

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4 years ago

Twice the difference between a number, x, and forty-four is three times the number

nexus9112 [7]

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4 years ago

Qyr = a1 + dan 1)

Ulleksa [173]

Answer:320

Step-by-step explanation:

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3 years ago

Verify the pythagorean identity 1 + c o t ^2 θ = c s c ^2 θ

Grace [21]

Answer: The identity is verified. See the explanation.

Step-by-step explanation:

You must keep on mind the following identities:

$csc\theta=\frac{cos\theta}{sin\theta}\\\\csc\theta=\frac{1}{sin\theta}\\\\sin^2\theta+cos^2\theta=1$

Therefore, by substitution, you can rewrite the identity as shown below:

$1+cot^2\theta=csc^2\theta\\\\1+\frac{cos^2\theta}{sin^2\theta}=csc^2\theta$

Simpliying, you obtain:

$\frac{sin^2\theta+cos^2\theta}{sin^2\theta}=csc^2\theta\\\\\frac{1}{sin^2\theta}=csc^2\theta\\\\csc^2\theta=csc^2\theta$

The identity is verified.

6 0

3 years ago

Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho

oksano4ka [1.4K]

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

7 0

3 years ago