2 years ago

If a=3/6 and b=-3/-5 find -3a+4b

Mathematics

1 answer:

Fiesta28 [93]2 years ago

7 0

Answer:

Step-by-step explanation:

-3(3/6) + 4(3/5)

-3(1/2) + 4(3/5)

-3/2 + 12/5

-15/10 + 24/10

9/10

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vlada-n [284]

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2 years ago

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$