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3 years ago

What type of polynomial is P(x) = 7x3 + 2x2 – 3x4?

Mathematics

1 answer:

gregori [183]3 years ago

5 0

Answer:

quartic

Step-by-step explanation:

p(x) = 7 x^3 + 2 x^2 - 3 x^4

The degree of an individual term of a polynomial is the exponent of its variable; the exponents of the terms of this polynomial are, in order, 3, 2, and 4.

The degree of the polynomial is the highest degree of any of the terms; in this case, it is 4.

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Graph the line with the equation y = 1/2x+3

Alecsey [184]

Answer:

(0,3) (-6,0) these will be the points

7 0

2 years ago

The sum of a number and 5 times a second number is represented by the equation a+5b=19 . The sum of half of the first number and

TiliK225 [7]

Here you have a system of 2 equations and 2 unknowns. An easy way to solve this type is to isolate one of the variables.

12a + 2b = 8
2b = 8 - 12a
b =
$b = \frac{8 - 12a }{2}$
b = 4 - 6a

Now plug 4 - 6a into equation 1 to solve for a.

a + 5(4 - 6a) = 19

a + 20 - 30a = 19

-29a = -1

a = 1/29 (answer)

Now plug a into the equation for b

b= 4 - 6(1/29)

b= 110/29 (answer)

4 0

3 years ago

Find the value of the derivative (if it exists) at the indicated extremum. (If an answer does not exist, enter DNE.)

Liono4ka [1.6K]

Answer:

The answer is "0".

Step-by-step explanation:

Please find the correct question in the attached file.

Given value:

$\to f(x) =\frac{x^2}{x^2+5}\\\\$

Formula:

$\bold{\frac{d}{dx} \frac{u}{v} = \frac{ v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}}$

$\to f'(x) =\frac{(x^2+5) 2x -2x^3 (2x)}{(x^2+5)^2}\\\\\to f'(0) =\frac{(0^2+5)2(0) -4(0)^4}{(0^2+5)^2}\\\\\to f'(0) =\frac{(0)(5) -2(0)}{(5)^2}\\\\\to f'(0) =\frac{0 - 0}{(5)^2}\\\\\to f'(0) =\frac{0}{25}\\\\\to f'(0) = 0$

6 0

2 years ago

– 2y – 10 = x<br> - 10x - y = -14

fredd [130]

Answer:

hey there guys ^^

Step-by-step explanation:

5 0

3 years ago

HELP TIMER Write the equation of a hyperbola centered at the origin with x-intercept +/- 4 and foci of +/-2(squareroot 5)

nikitadnepr [17]

Answer:

$\frac{x2}{a} - \frac{y2}{b2} = 1$

Step-by-step explanation:

A hyperbola is the locus of a point such that its distance from a point to two points (known as foci) is a positive constant.

The standard equation of a hyperbola centered at the origin with transverse on the x axis is given as:

$\frac{X2}{16} - \frac{b}{4} = 1$

The coordinates of the foci is at (±c, 0), where c² = a² + b²

Given that a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2√5. Since the x intercept is ±4, this means that at y = 0, x = 4. Substituting in the standard equation:

I don't feel like explaining so...

a. = 4

The foci c is at +/-2√5, using c² = a² + b²:

B = 2

Substituting the value of a and b to get the equation of the hyperbola:

$\frac{x2}{a2} - \frac{y2}{b2} = 1$

$\frac{x2}{16} - \frac{b2}{4} = 1$

4 0

2 years ago