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3 years ago

9

a gold bar that is 16 centimeters by 2.5 centimeters by 5 centimeters has a density of 19.3 grams per cubic centimeter. what is

the mass of the gold bar?

2 answers:

insens350 [35]3 years ago

8 0

The volume of the bar is:
16*2.5*5= 200 cm3
the mass is:
200*19.3= 3860 grams

Delvig [45]3 years ago

4 0

Answer:

Mass of the gold bar is 3860 grams.

Step-by-step explanation:

The dimensions of the gold bar is 16 centimeters by 2.5 centimeters by 5 centimeters.

The density of the gold bar is 19.3 grams per cubic centimeter.

We know the formula,

$\text{density}=\frac{\text{mass}}{\text{volume}}$

Volume = 16×2.5×5 = 200 cubic centimeters.

Substituting the known values in the above formula, we get

$19.3=\frac{\text{mass}}{200}\\\\\text{mass}=200\times19.3\\\\\text{mass}=3860\text{ grams}$

Hence, mass of the gold bar is 3860 grams.

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Identify the augmented matrix for the system of equations.

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Answer:

The augmented matrix for the system of equations is $\left[\begin{array}{cccc}0&2&-3&1\\7&0&5&8\\4&1&-3&6\end{array}\right]$ .

Step-by-step explanation:

This system consists in three equations with three variables ( $x$ , $y$ , $z$ ).The augmented matrix of a system of equations is formed by the coefficients and constants of the system of linear equations. In this case, we conclude that the system of equations has the following matrix:

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Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

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Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

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According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

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$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

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(b)

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The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

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Substitute x=2 in equation 2.

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The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

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$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

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