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grin007 [14]
3 years ago
11

The sum of two polynomials is –yz2 – 3z2 – 4y + 4. If one of the polynomials is y – 4yz2 – 3, what is the other polynomial? –2yz

2 – 4y + 7–2yz2 – 3y + 1–5yz2 + 3z2 – 3y + 13yz2 – 3z2 – 5y + 7
Mathematics
2 answers:
lidiya [134]3 years ago
4 0

Answer:

D. 3yz2 – 3z2 – 5y + 7

Step-by-step explanation:

marishachu [46]3 years ago
3 0

The first polynomial is y -4yz^2-3 and let the second polynomial be P(y,z).

If the sum of two polynomials is -yz^2-3z^2-4y+4, then y-4yz^2-3+P(y,z)=-yz^2-3z^2-4y+4.

Take the first polynomial from the left side to the right side:

P(y,z)=-yz^2-3z^2-4y+4 -(y-4yz^2-3),\\ P(y,z)=-yz^2-3z^2-4y+4 -y+4yz^2+3,\\ P(y,z)=3yz^2-3z^2-5y+7.

Answer: the second polynomial is 3yz^2-3z^2-5y+7.

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Answer: 10, 11, & 12

<u>Step-by-step explanation:</u>

Let x represent the age of the youngest child.  

Their ages are consecutive so,

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Middle: x + 1

Oldest: x + 2

The age of the Youngest squared (x²) equals 8 times the Oldest [8(x + 2)] plus 4.

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4 0
3 years ago
In a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of th
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Answer:

Required probability is 0.784 .

Step-by-step explanation:

We are given that in a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of the students passed both exams.

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Now, if the student passed the first exam, probability that he passed the second exam is given by the conditional probability of P(S/F) ;

As we know that conditional probability, P(A/B) = \frac{P(A\bigcap B)}{P(B) }

Similarly, P(S/F) = \frac{P(S\bigcap F)}{P(F) } = \frac{P(F\bigcap S)}{P(F) }  {As P(F \bigcap S) is same as P(S \bigcap F) }

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Therefore, probability that he passed the second exam is 0.784 .

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