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3 years ago

Anyone help? What is the derivative of x^2 + 2x - 4? Thanks!

2 answers:

8 0

D/dx(x² + 2x - 4) = 2x + 2

d/d (x² + 2x - 4)

= d/dx (x²) + d/dx(2x) - d/dx(4)

= d/dx(x^2) - d/dx(2x) = 2x (due to power rule)

Replug in to equation
d/dx (2x) = 2

Take the constant out

d/dx (x) = 1

apply the common derivative

2 x 1

simplify

2 x 1 = 2

d/dx(4) = 0

d/dx(4)

drivative of a constant = 0

= 2x + 2 - 0

Simplify:
2x + 2

2x + 2 is your answer

hope this helps

castortr0y [4]3 years ago

4 0

Let's take the derivative of each term one by 1

x^2
Apply the power rule for derivatives.
2x^(2-1) = 2x
You get 2x.

2x
Apply the power rule (x^1) and the constant multiple rule.
2 * x^(1-1) = 2 * x^0 = 2
You get 2.

-4
The derivative of a constant is 0.

Put that together and you get 2x + 2

Have an awesome day! :)

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1. Seekor burung helang berada 4.56 m di atas aras laut manakala seekor kuda laut berada 6.82 m

s2008m [1.1K]

(a) Calculate the vertical distance between the eagle and the seahorse.

Answer a: 11.38 meters is the vertical distance between them.

(b) Describe Adrian's position from the sea level.

Answer b: Adrian is 1.13m below the sea level.

Step-by-step explanation:

(a) Since sea level is the origin of coordinates. distances above sea level are positive and distances below sea level are negative. The distance between two points is the difference between them.

The vertical distance between the eagle and the seahorse is the difference between their positions:

Distance = 4.56m - (-6.82m) = 4.56m + 6.82m = 11.38 meters

Answer a) 11.38 meters is the vertical distance between them.

(b) Adrian's position is in the middle of the distance between the eagle and the seahorse:

Middle = 11.38m/2 = 5.69m

Since 5.69m is more than 4.56m above the sea level, then Adrian is below the sea level:

4.56m - 5.69m = -1.13m

Answer b: Adrian is 1.13m below the sea level.

<h2><em>Spymore</em></h2>

6 0

3 years ago

C.) 0,6 :3\5 = 4\9: ×<br>

Natalka [10]

Answer:

There is no question. Please provide a question.

7 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$