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3 years ago

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A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass

of the monument to the nearest tenth.

2 answers:

Sladkaya [172]3 years ago

8 0

Let units cancel out just like numbers...

25365.4cm^3(2.7g/cm^3)=68486.58g which is 68486.6g (to nearest tenth)

RideAnS [48]3 years ago

4 0

(2.7g/cm^3)(25365.4cm)=mass
=68486.58g
or, rounded, 68486.6grams

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$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

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2 years ago

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3 years ago

Read 2 more answers

Write an equation for an ellipse centered at the origin, which has foci at (0,\pm\sqrt{63})(0,± 63 )left parenthesis, 0, comma

steposvetlana [31]

Answer:

$\frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }$

Step-by-step explanation:

Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) $\frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} }$ .

We find the co-vertices b from b = ±√(a² - c²) where a = 91 and c = 63

b = ±√(a² - c²)

= ±√(91² - 63²)

= ±√(8281 - 3969)

= ±√4312

= ±14√22

So the equation is

$\frac{x^{2} }{(14\sqrt{22}) ^{2} } + \frac{y^{2} }{91^{2} } = \frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }$

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