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3 years ago

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3. A mysterious metal medallion has a mass of 3,088 grams. Water is poured into a graduated glass container with a 10-by-10 cm s

quare base with a water level of 53.0 cm. The medallion is dropped into the container and the water level rises to 54.6 cm. Find the volume and the density of the

1 answer:

velikii [3]3 years ago

6 0

Answer:

<u>Initial volume:</u>

10*10*53 = 5300 cm³

<u>Increased volume:</u>

10*10*54.6 = 5460 cm³

<u>Volume of the medallion:</u>

5460 - 5300 = 160 cm³

Density = mass / volume

<u>Density is:</u>

3.088/160 = 19.3 g / cm³

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PLEASE HELP. <br> I know how to do these but I'm stuck on this one.

Nezavi [6.7K]

Steve ran 11 miles

let the distance Steve ran be x

Then distance Kevin ran is x + 4 ( 4 miles more than Steve )

x + x + 4 = 26 ( the sum of their distance is 26 )

2x + 4 = 26 ( subtract 4 from both sides )

2x = 22 ( divide both sides by 2 )

x = 11

Steve ran 11 mies and Kevin ran 11 + 4 = 15 miles

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4 years ago

12. Si R(x)=-5x2+x-2 y S(x)=-x - 3x, se<br>¿cuál es el resultado de R(x) + S(x) ?

lesya692 [45]

Answer:

El resultado eres: $R(x) + S(x) = -6x^2 - 2x - 2$

Step-by-step explanation:

Se dan las siguientes funciones

$R(x) = -5x^2 + x - 2$

$S(x) = -x^2 - 3x$

Cuál es el resultado de R(x) + S(x) ?

Se soman los fatores en comun. Entonces:

$R(x) + S(x) = -5x^2 + x - 2 + (-x^2 - 3x) = -5x^2 + x - 2 - x^2 - 3x = -5x^2 - x^2 + x - 3x - 2 = -6x^2 - 2x - 2$

El resultado eres: $R(x) + S(x) = -6x^2 - 2x - 2$

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3 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

<u />

(b) The rate the area formed by the ladder is changing is approximately <u>-75.29 ft.²/sec</u>

<u />

(c) The rate at which the angle formed with the wall is changing is approximately <u>0.286 rad/sec</u>.

Reasons:

The given parameter are;

Length of the ladder, <em>l</em> = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let <em>y</em> represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ <u>-6.86 m/s downwards</u>

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ <u>-75.29 ft.²/sec</u>

(c) From trigonometric ratios, we have;

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ <u>0.286 rad/sec</u>

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago