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3 years ago

Please help me on this one?

Mathematics

2 answers:

RideAnS [48]3 years ago

4 0

It should be A because you are trying to solve for X

blondinia [14]3 years ago

3 0

3x + 4y = 16 Subtract 4y from both sides. Put brackets around the 2 terms on the right.

3x + 4y - 4y = (16 - 4y)

3x = (16 - 4y) Divide by 3

$\text{ x =}\dfrac{(16 - 4y)}{3} = \dfrac{16}{3} - \dfrac{4y}{3}$

Answer That's B.

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4 years ago

Translate the following sentence into an equation, then find the number. Five times a number less than 18 is is equal to 13.

Nata [24]

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The answer would be 5x-18=13

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7 0

4 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

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3 years ago