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3 years ago

What is the value of x in the equation -6x=5x+22? -22 O.-2 2 22

Mathematics

1 answer:

Ivanshal [37]3 years ago

8 0

so the right answer is -2.

Look at the attached picture

Hope it will help you....

Good luck on your assignment

~prãgyâ

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There are 16 ounces in one pound.

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2 years ago

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Two ships leave the same port in different directions, forming a 120° angle between them. One ship travels 70 mi. and the other

kondaur [170]

Answer : Distance between the ships to the nearest miles = 106.03 ≈ 106 mi.

Explanation :

Since we have shown in the figure below :

a=70 mi.

b=52 mi.

c=x mi.

$\text{Since two ships leaves the same port in different directions forming a }120\textdegree\text{angle between them.}$

So, we use the cosine rule , which states that

$c^2=a^2+b^2-2ab.cosC\\\\x^2=70^2+52^2-2\times 70\times 52\times cos(120\textdegree)\\\\x^2=4900+2704-7280\times (-0.5)\\\\x^2=7604+3640\\\\x^2=11244\\\\x=\sqrt{11244}\\\\x=106.03$

So, c = x= 106.03 mi.

Hence, distance between the ships to the nearest miles = 106.03 ≈ 106 mi.

8 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

2 years ago