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2 years ago

6

A rock weighs 80 g on a triple beam scale. The same rock is then gently placed inside a graduated cylinder containing 30 mL of w

ater. If the water level in the graduated cylinder rises to 75 mL. What is the density of the rock

1 answer:

Alexxandr [17]2 years ago

3 0

Answer:

1.78 g/cm³

Step-by-step explanation:

<u>80 g rock displaces water:</u>

75 ml - 30 ml = 45 ml = 45 cm³

<u>Density = mass / volume:</u>

d = 80 g / 45 cm³
d = 1.78 g/cm³ (rounded)

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Step-by-step explanation:

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When you draw an image in the right triangle, is it always reflected?

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Find a12 given the geometric sequence 4, 8, 16, 32, ...

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The <em>twelfth</em> element of the <em>geometric</em> sequence is equal to 4,096. (Correct choice: D)

<h3>How to find a determined element of a geometric sequence by exponential formulae</h3>

Sequences are series of elements generated according to at least one condition, usually equations. <em>geometric</em> sequences are generated according to a <em>exponential</em> formulas, whose form and characteristics are described below:

f(n) = a · bⁿ ⁻ ¹ (1)

Where:

a - First element of geometric sequence
b - Common ratio of the geometric sequence
n - Element index within the geometric sequence

If we know that a = 4, b = 2 and n = 12, then the twelfth element of the geometric sequence from the statement is:

f(12) = 4 · 2¹² ⁻ ¹

f(12) = 4 · 2¹¹

f(12) = 4 · 2,048

f(12) = 4,096

The <em>twelfth</em> element of the <em>geometric</em> sequence is equal to 4,096. (Correct choice: D)

To learn more on geometric sequences: brainly.com/question/4617980

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2 years ago

at a highschool, the length of a class period is 40 minutes. What is the length, in hours, of a class period at the high school?

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3 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

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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3 years ago

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