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2 years ago

Can someone please help me

Mathematics

1 answer:

Dennis_Churaev [7]2 years ago

3 0

Okay, so I had to resort to the calculator since I couldn't figure out how to do it, anyway here.

(-0.08928571428) -this is the answer according to it id says take a few numbers and see what answer fits it the most but yeah, I didn't help much.

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A = {a, b, c, d}

Helen [10]

Answer:

Johnny is wrong.

$A \cup B = \left\{a,b,c,d,e,f,g,h\right\}$

Step-by-step explanation:

Johnny is wrong.

A better definition would be: $A \cup B$ is the set of all elements that belong to at least one A or B.

So, the elements that belong to both A and B, like c and d in this exercise, also belong to $A \cup B$ .

So:

$A \cup B=\left\{a,b,c,d,e,f,g,h\right\}$

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3 years ago

Which situation can the integer -71 represent?

hoa [83]

The answers C because it is below sea level and sea level is 0 so it has to be negative

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2 years ago

Laura walked 612 laps of a circular path in 2 hours. At that rate, how many laps of the path did she walk in one hour?

kari74 [83]

Answer:

306 laps

Step-by-step explanation:

If Laura walked 612 laps in 2 hours then if you divide 2 by 2 then it will give you 1, which is 1 hour. This means that if you divide 612 by 2, it should give you the answer of 306, the amount of laps in 1 hour. I hope this helps!

8 0

2 years ago

Katie memorized 40 multiplication facts last week. This week she memorized 56 multiplication facts. What is her percent of incre

Oksi-84 [34.3K]

Answer:

40%

Step-by-step explanation:

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2 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)

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

5 0

2 years ago