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3 years ago

14

Solve 14n+6p-8b=18p for n

1 answer:

kherson [118]3 years ago

6 0

Answer:

hope this helps :)

Step-by-step explanation:

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How does an earthquake of magnitude 7.9 compare in intensity with an earthquake of magnitude of 3.7 on the Richter scale? The l

Andre45 [30]

We are required to compare the magnitude for two earthquake intensity:
the larger intensity is 7.9
the smaller intensity is 3.7
thus the ratio of the intensity will be:
(larger intensity)/(smaller intensity)
=7.9/3.7
=2.135
~2
Thus we conclude that:
The larger earthquake's intensity was 2 times as great as the smaller earthquake's intensity.

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3 years ago

What is the value of c?<br><br> ______ units

Shalnov [3]

The answer is 5, C = 5 units.

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2 years ago

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Whats 21/40 in equivalent decimals

GREYUIT [131]

The equivalent decimals for 21/40 is 0.525

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2 years ago

Find the area between the graph of the function and the x-axis over the given interval, if possible.

vodomira [7]

Answer:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

Step-by-step explanation:

In order to calculate the area between the function and the x axis we need to solve the following integral:

$A = \int_{-\infty}^1 \frac{5}{(x-2)^2}$

For this case we can use the following substitution $u = x-2$ and we have $dx = du$

$A = \int_{a}^b \frac{5}{u^2} du = 5\int_{a}^b u^{-2}du$

And if we solve the integral we got:

$A= -\frac{5}{u} \Big|_a^b$

And we can rewrite the expression again in terms of x and we got:

$A = -\frac{5}{x-2} \Big|_{-\infty}^1$

And we can solve this using the fundamental theorem of calculus like this:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

7 0

3 years ago

High grade steel consists of 85% iron and 15% magnese. Low grade steel consists of 67% iron and 33% mag ese. NASA orders 500 ton

Mamont248 [21]

Answer:

$361\dfrac{1}{9}$ tons of high steel and $138\dfrac{8}{9}$ tons of low steel

Step-by-step explanation:

Let x be the number of tons of high grade steel and y be the number of tons ow low grade steel needed.

In x tons of high grade steel there are

$0.85x$ tons of iron

$0.15x$ tons of magnese

In y tons of low grade steel there are

$0.67y$ tons of iron

$0.33y$ tons of magnese

NASA orders 500 tons of steel, so

$x+y=500$

and specifies that it must be in the proportion 80% iron and 20% magnese, so

$0.85x+0.67y=500\cdot 0.8\\ \\0.85x+0.67y=400$

From the first equation,

$x=500-y$

Substitute it into the second equation:

$0.85(500-y)+0.67y=400\\ \\425-0.85y+0.67y=400\\ \\-0.18y=-25\\ \\y=\dfrac{2,500}{18}=138\dfrac{8}{9}\ tons\\ \\x=361\dfrac{1}{9}\ tons$

7 0

3 years ago