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Makovka662 [10]

1 year ago

15

Solvex + 3 = −6.

1 answer:

Blizzard [7]1 year ago

4 0

Step-by-step explanation:

please mark me as brainlest

You might be interested in

If f(x) = 9x10 tan−1x, find f '(x).

djverab [1.8K]

Answer:

$\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle f(x) = 9x^{10} \tan^{-1}(x)$

<u>Step 2: Differentiate</u>

[Function] Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Basic Power Rule: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Arctrig Derivative: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

2 years ago

The Hu family goes out for lunch, and the price of the meal is $59. The sales tax on the meal is 7%, and the family also leaves

Setler79 [48]

$71.98 is the total cost for the meal
59.00
+ 4.13
+ 8.85
======
71.98

7 0

2 years ago

If ORBIT is coded BEOVG, then how is PLANET coded? /

arlik [135]

Answer:

1. CYNARG

Step-by-step explanation:

This a example of Caesar's Cipher, in which each letter in the original word leads to a ciphered letter according to the following equation:

$C = (P + o) \text{mod} 26$

In which C is the index of the Ciphered letter in the alphabet, P is the index of the original letter and o is the offset.

Finding the offset:

O is coded B

O is the 15th letter in the alphabet, so $P = 15$ .

B is the 2nd letter in the alphabet, so $C = 2$

$C = (P + o) \text{mod} 26$

$2 = (15 + o) \text{mod} 26$

$|o = |2-15| \text{mod} 26$

$o = 13$

So

$C = (P + 13) \text{mod} 26$

PLANET:

P

P is the 16th letter in the alphabet.

$C = (16 + 13) \text{mod} 26 = 3$

So P is coded C.

L

L is the 12th letter in the alphabet:

$C = (12 + 13) \text{mod} 26 = 25$

L is coded Y(25th letter in the alphabet)

A

A is the 1st letter in the alphabet

$C = (1 + 13) \text{mod} 26 = 14$

A is coded N

N

N is the 14th letter in the alphabet

$C = (14 + 13) \text{mod} 26 = 1$

N is coded A

E

E is the 5th letter in the alphabet

$C = (5 + 13) \text{mod} 26 = 18$

E is coded R

So the correct answer is:

1. CYNARG

4 0

2 years ago

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the ref

hichkok12 [17]

Answer:

P is exactly 3km east from the oil refinery.

Step-by-step explanation:

Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:

$\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}$

So the cost of laying pipe under water with this distance is

$800000\sqrt{(3 - d)^2 + 4}$

And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:

$800000\sqrt{(3 - d)^2 + 4} + 400000d$

We can find the minimum value of this by taking the 1st derivative and set it to 0

$800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0$

We can move the first term over to the right hand side and divide both sides by 400000

$1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}$

$\sqrt{(3 - d)^2 + 4} = 6 - 2d$

From here we can square up both sides

$(3 - d)^2 + 4 = (6 - 2d)^2$

$9 - 6d + d^2 + 4 = 36 - 24d + 4d^2$

$3d^2-18d+27 = 0$

$d^2 - 6d + 9 = 0$

$(d - 3)^2 = 0$

$d -3 = 0$

d = 3

So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.

3 0

2 years ago

For each of the following vector fields

olga nikolaevna [1]

(A)

$\dfrac{\partial f}{\partial x}=-16x+2y$

$\implies f(x,y)=-8x^2+2xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(B)

$\dfrac{\partial f}{\partial x}=-8y$

$\implies f(x,y)=-8xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x$

$\implies \dfrac{\mathrm dg}{\mathrm dy}=x$

But we assume $g(y)$ is a function of $y$ alone, so there is not potential function here.

(C)

$\dfrac{\partial f}{\partial x}=-8\sin y$

$\implies f(x,y)=-8x\sin y+g(x,y)$

$\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

4 0

2 years ago