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3 years ago

If both slopes are undefined are they parallel?

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

8 0

Answer:

If the slopes of the lines are both zero, the lines are horizontal and are parallel by definition. Since the slopes of vertical lines are undefined and not considered equal, vertical lines will not be considered.

I hope this helps you :)

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3 years ago

Find the second derivative of 2x^3-3y^2=8

Umnica [9.8K]

Answer:

$\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

Step-by-step explanation:

Find the first implicit derivative using implicit differentiation

$2x^3-3y^2=8\\\\6x^2-6y\frac{dy}{dx}=0\\ \\-6y\frac{dy}{dx}=-6x^2\\ \\\frac{dy}{dx}=\frac{x^2}{y}$

Use the substitution of dy/dx to find the second derivative (d²y/dx²)

$\frac{d^2y}{dx^2}=\frac{(y)(2x)-(x^2)(\frac{dy}{dx})}{y^2}\\ \\\frac{d^2y}{dx^2}=\frac{2xy-(x^2)(\frac{x^2}{y})}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{2xy-\frac{x^4}{y}}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

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3 years ago

The Collatz conjecture is one of the most famous unsolved mathematical problems, because it's so simple, you can explain it to a

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Hi, you've asked an incomplete question. However, I inferred you need a brief explanation about The Collatz conjecture.

Explanation:

Put simply, what the Collatz conjecture unsolved problem entails is that if any positive number is picked and it is:

An even number (eg 2, 4, 6,...), then if they are divided by 2, the new number gotten should undergo the same process (that is to be divided by 2), it is believed your calculation would finally end up at 1. For example, let's pick the number 6, (6÷2=4; repeating the process 4÷2=1)
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2 years ago

A hog producer is feeding soybean meal and corn to his pigs. He needs to know how many pounds he needs to feed of each in order

Mazyrski [523]

Answer:

Feed 1, Soybean meal required = $\dfrac{2000}{3} \text{ lbs.}$

Feed 2, Corn meal required = $\dfrac{4000}{3} \text{ lbs.}$

Step-by-step explanation:

Total feed is 1 ton i.e. 2000 lbs.

Let x be the amount of Feed 1 required.

Feed 1 has $45\%$ of protein.

$\text{Protein in Feed 1 = }x \times \dfrac{45}{100} ..... (1)$

Then, amount of Feed 2 required = $(2000 - x) \text{ lbs.}$

Feed 2 has $15\%$ of protein.

$\text{Protein in Feed 2 = }(2000 - x) \times \dfrac{15}{100} ..... (2)$

As per question, total protein required is $25\%$ of 2000 lbs .

Adding (1) and (2) and putting it equal to total protein required.

$\Rightarrow x \times \dfrac{45}{100} + (2000 - x) \times \dfrac{15}{100} = 2000 \times \dfrac{25}{100}\\\Rightarrow 45x + 30000 - 15x = 50000\\\Rightarrow 30x = 20000\\\Rightarrow x = \dfrac{2000}{3}$

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Feed 2 required = $2000 - \dfrac{2000}{3}\\$

$\Rightarrow \dfrac{4000}{3}\text{ lbs.}$

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3 years ago