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2 years ago

Find the degree of the polynomial

Mathematics

1 answer:

ikadub [295]2 years ago

3 0

Answer:

Step-by-step explanation:

1) 5a - 2b² + 1

Degree of the polynomial = 2

2) 4x²-3x³ + 2x + 12

-3x³ + 4x² + 2x + 12

Degree of the polynomial = 3

To find the degree of the polynomial, arrange it in descending as per the power of the variables.

The highest power of the variable is the degree of the polynomial.

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Find the average rate of change for the function f(x) =x^2 -6x+2 on the closed side interval (-5,2)

s344n2d4d5 [400]

Answer:

- 9

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

$\frac{f(b)-f(a)}{b-a}$

Here [a, b ] = [ - 5, 2 ], thus

f(b) = f(2) = 2² - 6(2) + 2 = 4 - 12 + 2 = - 6

f(a) = f(- 5) = (- 5)² - 6(- 5) + 2 = 25 + 30 + 2 = 57 , thus

average rate of change = $\frac{-6-57}{2-(-5)}$ = $\frac{-63}{7}$ = - 9

6 0

2 years ago

Although 330° is a special angle on the unit circle, Amar wanted to determine its coordinates using the sum and difference formu

Dvinal [7]

Answer:

Here you go.Hope this help!!

8 0

2 years ago

Is -3/5 rational or irrational???????

GenaCL600 [577]

-3/5 is rational number

5 0

2 years ago

Read 2 more answers

Pemfaktoran dari 2x²+5x-18<br><br>

Harrizon [31]

We can factor a polynomial by finding its roots. In particular, a quadratic equation has (at most) two roots $x_1,\ x_2$ , which would allow us to write the polynomial as

$p(x)=a(x-x_1)(x-x_2)$

To find the solutions, we can use the quadratic formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{-5\pm\sqrt{169}}{4} = \dfrac{-5\pm13}{4}$

So, the two solutions are

$\dfrac{-5+13}{4} = 2\quad\text{and}\quad\dfrac{-5-13}{4} =-\dfrac{9}{2}$

And so we can factor the polynomial as follows:

$2x^2+5x-18=2(x-2)\left(x+\dfrac{9}{2}\right)$

3 0

2 years ago

Prove that the curve a(t) = (cost, sin 2t, cos 2t) is regular on R and that it self-intersects at (1,0,1). Check the self-inters

alexandr402 [8]

Answer:

The function a (t) is a vector function composed of the component functions $a_ {1} (t) = cost, a_ {2} (t) = sin2t$ and $a_ {3} (t) = cos2t$ . How $a_ {1} (t), a_ {2} (t), a_ {3} (t)$ are infinitely derivable functions in R, so they are regular functions in R.

Now, for $t = 0$ , you have to $a (0) = (cos (0), sin2 (0), cos2 (0)) = (1, 0, 1)$ . How the functions $a_ {1} (t), a_ {2} (t), a_ {3} (t)$ are periodic functions with period $2 \pi,$ the vector function $a (t)$ will take the same point $(1, 0 , 1)$ at $t = 2n\pi, n = 0, 1, 2, 3, ...$ then the vector function is auto-intercepted

Step-by-step explanation:

6 0

2 years ago