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2 years ago

Hii guys please answer this question

Mathematics

1 answer:

Mariana [72]2 years ago

5 0

Answer:

-1/2

Step-by-step explanation:

Zeros of a polynomial function are its x-intercepts, the values of x that make the polynomial zero. Each zero x=p corresponds to a linear factor (x-p) of the polynomial.

<h3>Monic polynomial</h3>

A monic polynomial is one that has a leading coefficient of 1. For a monic polynomial of degree n, the coefficient of the term of degree n-1 is the opposite of the sum of all of the zeros. The constant term of any odd-degree monic polynomial is the opposite of the product of the zeros.

The given polynomial can be made monic by dividing by its leading coefficient:

(2x^3 -x^2 -5x -2)/2 = x^3 -(1/2)x^2 -(5/2)x -1

<h3>Sum of zeros</h3>

The given polynomial has degree 3, so we're interested in the coefficient of the x^2 term. That coefficient is -1/2, so the sum of zeros will be ...

sum of zeros = -(-1/2) = 1/2

If z represents the third zero, then we have the sum ...

1/2 = -1 +2 +z

z = -1/2 . . . . . . subtract 1 from both sides

The third zero is -1/2.

<h3>Product of zeros</h3>

The given polynomial is of odd degree, so the product of the zeros is the opposite of the constant.

(-1)(2)(z) = -(-1)

z = 1/(-2) = -1/2 . . . . . divide by the coefficient of z

The third zero is -1/2.

<h3>Graph</h3>

Of course, a graph will show all of the (real) zeros. The attached graph from a graphing calculator shows the third zero is -1/2.

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elena55 [62]

Answer:

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Step-by-step explanation:

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3 years ago

The length of a rectangle is increasing at a rate of 1 meter per day and the width is decreasing at a rate of 2 meters per day.

Goshia [24]

Answer:

The area is changing at 2 m²/d.

Step-by-step explanation:

The area of a rectangle is given by:

$A = l*w$

Where:

l is the length

w is the width

We have that the length is increasing at a rate:

$\frac{dl}{dt} = 1 \frac{m}{d}$

And the width is decreasing at a rate:

$\frac{dw}{dt} = -2 \frac{m}{d}$

The change in the rectangle's area is the following:

$\frac{dA}{dt} = w\frac{dl}{dt} + l\frac{dw}{dt}$

When the length is 8 meters and the width is 18 meters we have:

$\frac{dA}{dt} = 18 m*1\frac{m}{d} + 8 m(-2 \frac{m}{d})$

$\frac{dA}{dt} = 18 \frac{m^{2}}{d} - 16 \frac{m^{2}}{d}$

$\frac{dA}{dt} = 2 \frac{m^{2}}{d}$

Therefore, the area is changing at 2 m² per day.

I hope it helps you!

8 0

3 years ago

A cylinder has diameter of 4ft and a height of 9ft. Explain whether halving the diameter has the same effect on the surface area

elena55 [62]

Answer:

See Below

Step-by-step explanation:

The surface area of cylinder is given by the formula:

$SA=2\pi r^2 + 2\pi r h$

Where

r is radius ( diameter is 4, so radius is 4/2 = 2)

h is height ( h = 9)

Lets find original surface are:

$SA=2\pi r^2 + 2\pi r h\\SA=2\pi (2)^2 + 2\pi (2) (9)\\SA=8\pi +36\pi\\SA=44\pi$

<u>Halving diameter:</u>

diameter would be 4/2 = 2, so radius would be 2/2 = 1

So, SA would be:

$SA=2\pi r^2 + 2\pi r h\\SA=2\pi (1)^2 + 2\pi (1) (9)\\SA=2\pi +18\pi\\SA=20\pi$

<u>Halving height:</u>

Height is 9, halving would make it 9/2 = 4.5

Now, calculating new SA:

$SA=2\pi r^2 + 2\pi r h\\SA=2\pi (2)^2 + 2\pi (2) (4.5)\\SA=8\pi + 18\pi\\SA= 26\pi$

Original SA is $44\pi$ ,

Halving diameter makes it $20\pi$

Halving height makes it $26\pi$

So, halving diameter does not have same effect as halving height.

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