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3 years ago

In the equation 3x+3y+3z=18 what does x equal

1 answer:

5 0

<span>3x+3y+3z=18

Start by subtracting both sides by 3z

3x + 3y = 18 - 3z

Now , you can subtract 3y from both sides....

3x = - 3z - 3y + 18

Finally .... To get x by itself divide both sides by 3.......

x = - z - y + 6 is the answer</span>

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Can anyone solve this correctly?

andrew-mc [135]

Answer:

A. $200.96 \: {ft}^{2}$

Step-by-step explanation:

Corral is circular in shape.

Radius of the corral would be be the distance from the post to the edge of the corral.

-> r = 8 ft

$A(grass\: eaten) = \pi {r}^{2} \\ \\ = 3.14 {(8)}^{2} \\ \\ = 200.96 \: {ft}^{2}$

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2 years ago

Santo made 4 less than twice as many cupcakes as Lynn. Write an equation to represent the number of cupcakes that Santo made.

pickupchik [31]

Santo made 4 less than twice as many cupcakes as Lynn.

x * 2 - 4

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3 years ago

3. Which of the following are the roots of the quadratic function below?

sergejj [24]

Answer:

x = ± 12

Step-by-step explanation:

Given

f(x) = x² - 144

To find the roots set f(x) = 0, that is

x² - 144 = 0 ( add 144 to both sides )

x² = 144 ( take the square root of both sides )

x = ± $\sqrt{144}$ = ± 12

4 0

2 years ago

Read 2 more answers

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

Explain how the sum and difference formula cos(A+/- B) = cosA cosB -/+ sinA sinB could be useful?

serg [7]

One way it's useful is in the application of phasors in physics. Phasors require you to add up vectors of two different angles, so the cosine angle addition formula can be used if you already know the cosine and sine of the original vector angles.

6 0

3 years ago

Read 2 more answers