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3 years ago

What is the quotient? 2 1/2÷(−1 3/4)

Mathematics

1 answer:

anastassius [24]3 years ago

5 0

Answer:

$2 \times \frac{1}{2} \div ( - \frac{13}{4}) = \frac{5}{2} \times - \frac{4}{3} = - \frac{10}{13}$

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Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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2 years ago

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2 years ago

The expression (x − 4)2 is equivalent to x2 − 16 x2 + 16 x2 − 8x + 16 x2 + 8x + 16

zzz [600]

Answer:

Given below

Step-by-step explanation:

The algebraic identity is (a-b)^2

= a^2 + b^2 - 2ab

So it'll be,

(x-4)^2

= x^2 + (4)^2 - (2)(x)(4)

= x^2 -8x + 16

or x^2 +16 -8x

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3 years ago

A rectangle is placed around a semi circle as shown below. The width of the rectangle is 7yd. find the area of the shaded region

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Answer:

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Step-by-step explanation:

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The area of the shaded region is the same as the area of the semicircle subtracted from the area of the rectangle.

area of shaded region = area of rectangle - area of semicircle

A = LW - (1/2)(pi)r^2

A = 14 yd * 7 yd - (1/2)(3.14)(7 yd)^2

A = 98 yd^2 - (1.57)(49 yd^2)

A = 98 yd^2 - 76.93 yd^2

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3 years ago

Please help! Correct answer only, please! The following information matrices shows how many of each vehicle type sold and the bo

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Answer: B) Scott sold 1 van

<u>Step-by-step explanation:</u>

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3 years ago