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3 years ago

What can you conclude about parallel lines drawn in the coordinate plane?

Mathematics

1 answer:

anygoal [31]3 years ago

6 0

Parallel lines are both vertical and they don’t meet.

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Gelneren [198K]

Answer:

Im pretty sure the answe is C

Step-by-step explanation:

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3 years ago

Which of the following quantities are to the left of zero on a number line? Select three that apply.

emmainna [20.7K]

Answer:

A, D, and E

Step-by-step explanation:

because they are negative

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3 years ago

What kind of angle is formed by the tangent to a circle and a radius of that circle?

Vsevolod [243]

A tangent is always perpendicular to the radius at the point of tangency.

A right angle.

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3 years ago

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What is -2(3x+12y-5-17x-16y+4) simplified?

CaHeK987 [17]

28x + 8y + 2 is the simplified version.

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3 years ago

Read 2 more answers

The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Th

Fantom [35]

Answer:

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C$

Step-by-step explanation:

To find:

∫ $6x^5(x^6-2)\,dx$

Solution:

Method of substitution:

Let $x^6-2=t$

Differentiate both sides with respect to $t$

$6x^5\,dx=dt$

[use $(x^n)'=nx^{n-1}$ ]

So,

∫ $6x^5(x^6-2)\,dx$ = ∫ $t\,dt$ = $\frac{t^2}{2}+C_1$ where $C_1$ is a variable.

(Use ∫ $t^n\,dt=\frac{t^{n+1} }{n+1}$ )

Put $t=x^6-2$

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1$

Use $(a-b)^2=a^2+b^2-2ab$

So,

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C$

where $C=2+C_1$

Without using substitution:

∫ $6x^5(x^6-2)\,dx$ = ∫ $6x^{11}-12x^5\,dx$ = $\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C$

So, same answer is obtained in both the cases.

7 0

2 years ago