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2 years ago

A solid silver spoon has a mass of 65.1g.

Mathematics

1 answer:

Keith_Richards [23]2 years ago

3 0

Answer:

D=M/V

D=65.1/6.2

Density is 10.5 grams per centimetre cube

Step-by-step explanation:

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Ruth wants to purchase a new video game that costs $59.00 and several bags of chips that sell for $1.99 each. She can spend no m

allsm [11]

Answer:

1.99c + 59 ≥ 75, where c ≥ 8.04

Step-by-step explanation:

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2 years ago

Add fractions and simplify the answer 1/2 + 2/3

IrinaVladis [17]

$\frac{1}{2} +\frac{2}{3} =\\\\ \frac{1\times 3}{2\times3} + \frac{2\times2}{3\times2} =\\\\\frac{3}{6} +\frac{4}{6}=\\\\\frac{7}{6} =\\\\ 1\frac{1}{6}\\\\ \frac{1}{2} +\frac{2}{3} =\boxed{\bf{\frac{7}{6} = 1\frac{1}{6}}}$

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2 years ago

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What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .

VladimirAG [237]

Answer:

the positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

$\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1$

Using the standard form of the equation:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1$

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term indicating that the hyperbola opens up and down.

a = distance that indicates how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

$y = \pm \frac{a}{b} (x-h)+k$

where;

$\frac{a}{b}$ is the slope

From above;

(h,k) = 11, 100

$a^2$ = 100

a = $\sqrt{100}$

a = 10

$b^2 =4$

b = $\sqrt{4}$

b = 2

$y = \pm \frac{10}{2} (x-11)+2$

$y = \pm 5 (x-11)+2$

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have

$\frac{a}{b}$ to be the slope in the equation $y = \pm \frac{10}{2} (y-11)+2$

$\frac{a}{b}$ = $\frac{10}{2}$

$\frac{a}{b}$ = 5

Thus, the positive slope of the asymptote = 5

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2 years ago

(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe

Fofino [41]

Answer:

a) $\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b) See Below for proper explanation

Step-by-step explanation:

a) The objective here is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is $e^x + 3 \ cos \ x$

The expansion is of $e^x$ is $e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...$

The expansion of cos x is $cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...$

Therefore; $e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]$