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3 years ago

Two fourth-degree polynomials are multiplied together. what is the degree of the product?

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

6 0

Answer:

The product of a fourth degree polynomial and a third degree polynomial is a 7th degree polynomial (just add the two degrees). Why is this the case? Remember, when you multiply variables with common bases, you add the exponents. Since the degree of a polynomial is just the largest exponent, you're really just adding the degrees when you multiply

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X^2 = -8x - 7 Rewrite the equation by completing the square. What are the solutions?

gladu [14]

1) rewrite the equation by completing the square

(X+4)^2=9

2) what are the solution to the equation

Answer A= X=-4 plus and minus 3

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3 years ago

Which of the following polynomials have degree 5?

saw5 [17]

Answer:

lll

Step-by-step explanation:

*dab

it feels good to know the answer

5 0

3 years ago

Read 2 more answers

A physical education teacher wanted to test whether her county’s students are falling below the national proportion. It is known

Ipatiy [6.2K]

Answer:

(a) The population parameter is students who ran the mile in less than 10 minutes.

(b) The hypothesis being tested is:

H0: p = 0.70

Ha: p < 0.70

(d) In order to conduct a one-sample proportion z-test, the following conditions should be met:

The data are a simple random sample from the population of interest.

The population is at least 10 times as large as the sample.

n⋅p ≥ 10 and n⋅(1−p) ≥ 10, where n is the sample size and p is the true population proportion.

n⋅p = 756 * 0.70 ≥ 10

n⋅(1−p) = 756 * (1 - 0.70) ≥ 10

All the conditions are met.

(e) p = 504/756 = 0.67

The test statistic, z = (p - p)/√p(1-p)/n = (0.67 - 0.70)/√0.70(1-0.70)/756 = -2.00

(f) The p-value is 0.0228.

(g) Since the p-value (0.0228) is less than the significance level (0.05), we can reject the null hypothesis.

(h) Therefore, we can conclude that less than 70% of students run one mile in less than 10 minutes.

Step-by-step explanation:

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3 years ago

Why can't a trapezoid have 4 obtuse angles

In-s [12.5K]

Hello there, and thank you for posting your question here on brainly.

A trapezoid is not possible to have for obtuse because it would look more like a diamond. Also, if it had 4 obtuse angles, it would have 6 sides.

Hope this helped!! ☺♥

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3 years ago

Read 2 more answers

Can someone help me do part two please? It’s very important send a picture or something. I don’t even care if you tell me the st

Nataly_w [17]

<h3>Explanation:</h3>

1. "Create your own circle on a complex plane."

The equation of a circle in the complex plane can be written a number of ways. For center c (a complex number) and radius r (a positive real number), one formula is ...

|z-c| = r

If we let c = 2+i and r = 5, the equation becomes ...

|z -(2+i)| = 5

For z = x + yi and |z| = √(x² +y²), this equation is equivalent to the Cartesian coordinate equation ...

(x -2)² +(y -1)² = 5²

2. "Choose two end points of a diameter to prove the diameter and radius of the circle."

We don't know what "prove the diameter and radius" means. We can show that the chosen end points z₁ and z₂ are 10 units apart, and their midpoint is the center of the circle c.

For the end points of a diameter, we choose ...

z₁ = 5 +5i
z₂ = -1 -3i

The distance between these is ...

|z₂ -z₁| = |(-1-5) +(-3-5)i| = |-6 -8i|

= √((-6)² +(-8)²) = √100

|z₂ -z₁| = 10 . . . . . . the diameter of a circle of radius 5

The midpoint of these two point should be the center of the circle.

(z₁ +z₂)/2 = ((5 -1) +(5 -3)i)/2 = (4 +2i)/2 = 2 +i

(z₁ +z₂)/2 = c . . . . . the center of the circle is the midpoint of the diameter

__₁₂₃₄

3. "Show how to determine the center of the circle."

As with any circle, the center is the <em>midpoint of any diameter</em> (demonstrated in question 2). It is also the point of intersection of the perpendicular bisectors of any chords, and it is equidistant from any points on the circle.

Any of these relations can be used to find the circle center, depending on the information you start with.

As an example. we can choose another point we know to be on the circle:

z₄ = 6-2i

Using this point and the z₁ and z₂ above, we can write three equations in the "unknown" circle center (a +bi):

|z₁ - (a+bi)| = r
|z₂ - (a+bi)| = r
|z₄ - (a+bi)| = r

Using the formula for the square of the magnitude of a complex number, this becomes ...

(5-a)² +(5-b)² = r² = 25 -10a +a² +25 -10b +b²

(-1-a)² +(-3-b)² = r² = 1 +2a +a² +9 +6b +b²

(6-a)² +(-2-b)² = r² = 36 -12a +a² +4 +4b +b²

Subtracting the first two equations from the third gives two linear equations in a and b:

11 -2a -21 +14b = 0

35 -14a -5 -2b = 0

Rearranging these to standard form, we get

a -7b = -5

7a +b = 15

Solving these by your favorite method gives ...

a +bi = 2 +i = c . . . . the center of the circle

4. "Choose two points, one on the circle and the other not on the circle. Show, mathematically, how to determine whether or not the point is on the circle."

The points we choose are ...

z₃ = 3 -2i
z₄ = 6 -2i

We can show whether or not these are on the circle by seeing if they satisfy the equation of the circle.

|z -c| = 5

For z₃: |(3 -2i) -(2 +i)| = √((3-2)² +(-2-i)²) = √(1+9) = √10 ≠ 5 . . . NOT on circle

For z₄: |(6 -2i) -(2 +i)| = √((6 -2)² +(2 -i)²) = √(16 +9) = √25 = 5 . . . IS on circle

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3 years ago