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3 years ago

Solve for n; 1.2n + 0.68 = 5

Mathematics

2 answers:

bulgar [2K]3 years ago

8 0

Answer:

n= 18 over 5

Step-by-step explanation:

kifflom [539]3 years ago

6 0

Answer: n = 3.6

Step-by-step explanation: When solving an equation that contains decimals, we want to get rid of the decimals as soon as possible.

Notice that 1.2 is taken to the tenths place

but 0.68 is taken to the hundredths place.

To get rid of all decimals in the equation, multiply both

sides by the greatest power of 10 that is required.

So here we multiply both sides by 100 which

moves all decimals 2 places to the right.

So we have (100)(1.2n + 0.68) = (5)(100).

This gives us 120n + 68 = 500.

Now subtract 68 from both sides to get 120n = 432.

Dividing both sides by 120, we find that n = 3.6.

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inysia [295]

Let's start by understanding that the graph is a linear line. We have a common difference if 4.5 for every 10x we have more.

We can start by finding the gradient, using rise over run. Let's take (10, 4.5) and (20, 9)

Our rise is 9 - 4.5 = 4.5 and our run is 10. Then, our gradient becomes 4.5/10 = 0.45

Now, we can substitute points using the point-slope form.

y - 4.5 = 0.45(x - 10)
y - 4.5 = 0.45x - 4.5
Hence, our line becomes y = 0.45x and you can verify by substituting in points from the table.

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3 years ago

Brainliest please help!

Dmitriy789 [7]

Answer:

120

Step-by-step explanation:

6 x 6 = 36

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3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

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3 years ago

A 56 yard loss in football game

o-na [289]

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3 years ago

if revenue for the week is 2000 and labor cost consist of two workers earning $8 per hour who work 40 hours each week what is la

DerKrebs [107]

The Answer is 32%.
Because for two workers for 40 hours each, it will cost 640
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3 years ago