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3 years ago

(7, 4) and (7,-2) solve?

Mathematics

1 answer:

scoundrel [369]3 years ago

7 0

Answer:Exact Form:

Decimal Form:

1.75

Mixed Number Form:

1 3/4

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Find a third-degree polynomial equation with rational coefficients that has roots –5 and 6 + i.

Vitek1552 [10]

The coefficients of the polynomial are rational, which means that any non-real roots occur alongside their complex conjugates. In this case, 6+i is a root, so 6-i is also a root.

So the simplest polynomial you can build with these roots is

(x - (-5)) (x - (6 + i )) (x - (6 - i )) = x ^3 - 7x ^2 - 23x + 185

(first choice)

8 0

3 years ago

Two lines that lie in parallel planes are

Alchen [17]

Answer: Parallel.

Two or more lines that lie in the same plane and never intersect. Parallel lines will always have the same slope. Skew lines are lines that are in different planes and never intersect. A transversal is a line that intersects two other lines.

6 0

3 years ago

Steve stayed after school for extra activities 55% of the 180 school days this year. How

Liula [17]

Answer:

He stayed for 99 days

Step-by-step explanation:

55% of 180 is 99

8 0

3 years ago

rooks are randomly placed on different squares of a chessboard. A rook is said to attack all of the squares in its row and its c

Olegator [25]

Answer: $2*8^{8}-8!$

Step-by-step explanation:

We can place a rook in each row in $8^{8}$ ways.

We can place a rook in each column in $8^{8}$ ways.

So the total ways in which we can place a rook in a row or column is $2*8^{8}$

Now there are 8 ways to choose the column for the first row, 7 ways to choose the column for the second row, and so on. So there are 8! ways to choose a column for a row.

So, we get the final answer by subtracting the 8! from the total ways a rook can be placed which is $2*8^{8}-8!$ when binomial coefficients are not evaluated

7 0

4 years ago

Prof Liu gave the same quiz to the students in her morning class and in her afternoon class. the average score for the two class

Paraphin [41]

Answer:

The number of students is more in afternoon class.

Step-by-step explanation:

The average marks for the two classes combined = 84

Average marks for the morning class, avg1 = 80

Average marks for afternoon class, avg2 = 86

The final average marks would be shifted more towards the average marks of the class with more number of students.

This is because the formula for final mean is

= $\frac{avg1 \times number of morning students + avg2 \times number of afternoon students}{total number of students}$

Since final mean 84 is closer to average of class 2 which is 86, hence the number of students is more in afternoon class.

3 0

3 years ago