All categories

3 years ago

It takes 30 minutes for a

Mathematics

1 answer:

CaHeK987 [17]3 years ago

4 0

Answer:

i believe 15 minutes

Step-by-step explanation:

You might be interested in

Mark and Julio are selling flower bulbs for a school fundraiser. Customers can buy bags of windflower bulbs and packages of croc

ikadub [295]

Answer:

Bag of windflower bulbs costs $8.50

Package of crocus bulbs costs $17.60

Step-by-step explanation:

Let $x be the price of one bag of windflower bulbs and $y be the price of one package of crocus bulbs.

1. Mark sold 2 bags of windflower bulbs for $2x and 5 packages of crocus bulbs for $5y. In total he earned $(2x+5y) that is $105. So,

2x+5y=105

2. Julio sold 9 bags of windflower bulbs for $9x and 5 packages of crocus bulbs for $5y. In total he earned $(9x+5y) that is $164.50. So,

9x+5y=164.50

3. You get the system of two equations:

$\left\{\begin{array}{l}2x+5y=105\\ \\9x+5y=164.50\end{array}\right.$

From the first equation

$5y=105-2x$

Substitute it into the second equation:

9x+105-2x=164.50

7x=164.50-105

7x=59.5

x=$8.50

So,

5y=105-2·8.5

5y=105-17

5y=88

y=$17.60

5 0

4 years ago

Find the derivative of following function.

Aleks04 [339]

Answer:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}$

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
$\displaystyle (uv)' = u'v + uv'$

Derivative Rule [Quotient Rule]:
$\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

Since the answering box is way too small for this problem, there will be limited explanation.

Step 1: Define

Identify.

$\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}$

Step 2: Differentiate

We can differentiate this function with the use of the given derivative rules and properties.

Applying Product Rule:

$\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '$

Differentiating the first portion using Quotient Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Differentiating the second portion using Quotient Rule again:

$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule again:
$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Substitute in derivatives:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Simplify:

$\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}$

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

∴ we have found our derivative.

---

Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

---

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

Unit: Differentiation

8 0

3 years ago

Read 2 more answers

Find the greatest common factor of the following monomials: 20a 46a^2 48a^6

Ierofanga [76]

60a^5 is the answer to your question

6 0

3 years ago

How do you math please help me

Stella [2.4K]

Math is math math gives more math math is math math is more math more math is tons of math tons of math is big brain math knowledge big brain math knowledge means your math

3 0

3 years ago

If x+y=5,x-y=4,find the value of x²-y²

inysia [295]

Answer:

Step-by-step explanation:

→ First find the value of x and y

x + y = 5

x - y = 4

→ Add the equations to cancel out the y's

2x = 9

→ Divide both sides by 2 to find the value of x

x = 4.5

→ Substitute x = 4.5 back into x - y = 4 to find the value of y

4.5 - y = 4

→ Minus 4.5 from both sides to isolate -y

-y = -0.5

→ Multiply everything by -1

y = 0.5

→ Substitute x = 4.5 and y = 0.5 into x² - y²

4.5² - 0.5² = 20.25 - 0.25 = 20

4 0

3 years ago