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2 years ago

Can someone help with this please

Mathematics

1 answer:

grigory [225]2 years ago

4 0

Polynomials are mathematical expressions involving variables raised with non-negative integers and coefficients. The polynomial is x³- 7x² - 6x + 72.

<h3>What is a polynomial?</h3>

Polynomials are mathematical expressions involving variables raised with non-negative integers and coefficients(constants who are in multiplication with those variables) and constants with only operations of addition, subtraction, multiplication and non-negative exponentiation of variables involved.

The polynomial using the given information can be made as shown below.

Given the zeros of the polynomial are -3, 4, and 6.

P(x) = (x+3)(x-4)(x-6) = x³- 7x² - 6x + 72

To check the end behaviour of the function, the graph of the function can be plotted as shown in the below image.

Hence, the polynomial is x³- 7x² - 6x + 72.

Learn more about Polynomials:

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find the equation of a circle which passes through the point (2,-2) and (3,4) and whose centre lies on the line x+y=2

Nadusha1986 [10]

Answer:

Equation of the circle

(x - 0.7)² + (y - 1.3)² = 12.58

Step-by-step explanation:

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle.

a) We are told in the question that the equation of the circle passes through point(2, -2)

Hence,

Substituting 2 for x and -2 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(2 - a)² +(-2 - b)² = r²

Expanding the bracket

(2 - a) (2 - a) + (-2 - b)(-2 - b) = r²

4 - 2a - 2a +a² +4 +2b +2b +b² = r²

4 - 4a + a² + 4 + 4b + b² = r²

a² + b² -4a + 4b + 4 + 4 = r²

a² + b² -4a + 4b + 8 = r²............Equation 1

We are also told that the equation of the circle also passes through point (3,4) also, where 3 = x and 4 = y

Hence,

Substituting 3 for x and 4 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(3 - a)² +(4 - b)² = r²

Expanding the bracket

(3 - a) (3 - a) + (4 - b)(4- b) = r²

9 - 3a - 3a +a² +16 -4b -4b +b² = r²

9 -6a + a² + 16 -8b + b² = r²

a² + b² -6a -8b + 9 + 16 = r²

a² + b² -6a -8b + 25 = r²..........Equation 2

The next step would be to subtract Equation 1 from Equation 2

a² + b² -4a + 4b + 8 - (a² + b² -6a -8b + 25) = r² - r²

a² + b² -4a + 4b + 8 - a² - b² +6a +8b - -25= r² - r²

Collecting like terms

a² - a² + b² - b² - 4a + 6a + 4b + 8b +8- 25 = 0

2a + 12b -17 = 0

2a + 12b = 17...........Equation 3

Step 2

We are going to have to find the values of a and b in other to get our equation of the circle.

Since the center of the circle(a, b) lies on x + y = 2

Therefore, we have

a + b = 2

a = 2 - b

2a + 12b = 17 ..........Equation 3

Substituting 2 - b for a in

2(2 - b) + 12b = 17

4 - 2b + 12b = 17

4 + 10b = 17

10b = 17 - 4

10b = 13

b = 13/10

b = 1.3

Substituting 1.3 for b in

a + b = 2

a + 1.3 = 2

a = 2 - 1.3

a = 0.7

hence, a = 0.7, b = 1.3

Step 3

We have to find the value of r using points (2, -2)

(x - a)² + (y - b)² = r²

Where x = 2 and y = -2

(-2 - 0.7)² + (-2 - 1.3)² = r²

(-2.7)² + (-3.3)² = r²

1.69 + 10.89 = r²

r² = 12.58

r = √12.58 = 3.55

Step 4

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle

a = 0.7

b = 1.3

r² = 12.58

Equation of the circle =

(x - 0.7)² + (y - 1.3)² = 12.58

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Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 1

Explanation:

The geometric mean can be represented by $\sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}$ .

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

$a_{n} = a_{1} • r^{n-1}$

Where $a_{n}$ is the nth term, $a_{1}$ is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, $a_{1}$ will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

$\sqrt[5]{\frac{a_{6}}{a_{1}}}$ =

$\sqrt[5]{\frac{1}{32}}$ =

$\frac{1}{2}$ .

Therefore: $r = \frac{1}{2}$ .

Using all the information we have, we can find the explicit rule:

$a_{n} = a_{1} • r^{n-1}$

$a_{1}$ = 32
$r = \frac{1}{2}$

$a_{n} = a_{1} • r^{n-1}$ →

$\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}$

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We can test that this works by substituting the number location of the term you want to find.

For instance:

$a_{1} = 32 • (\frac{1}{2})^{1-1}$

$a_{1} = 32 • (\frac{1}{2})^{0}$

$a_{1} = 32 • 1$

$a_{1} = 32$

$a_{6} = 32 • (\frac{1}{2})^{6-1}$

$a_{6} = 32 • (\frac{1}{2})^{5}$

$a_{6} = 32 • \frac{1}{32}$

$a_{6} = 1$

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