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3 years ago

What is the slope of tne line to 4x+6y=12

2 answers:

8 0

I don’t understand the other comment but do this
1)4x+6y=12 and subtract 4x from both sides
2) 6y= 12-4x. Then divide each side by 6
3) y= 2 -4/6 x.
-4/6 x is your slope. So simplify it to -2/3 and that’s your answer

RoseWind [281]3 years ago

3 0

The slope of 4x plus 6y equals 12 is x =3-3/2y

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Find the difference:

VLD [36.1K]

Answer: The difference cannot be found because the indices of the radicals are not the same.

Step-by-step explanation:

To find the difference you need to subtract the radicals. But it is important ot remember the following: To make the subtraction of radicals, the indices and the radicand must be the same.

In this case you have these radicals:

$\sqrt[ {8ab}^{3} ]{{ac}^{2} }- \sqrt[ {14ab}^{3}]{{ac}^{2} }$

You can observe that the radicands are the same, but their indices are not the same.

Therefore, since the indices are different you cannot subtract these radicals.

6 0

3 years ago

The bottle is completely full at 11:40 a.m. Which statements help explain this result? Check all that apply. The bottle was 1/2

Delicious77 [7]

Answer:

Step-by-step explanation:

Given that bottle was half full at 11.35 a.m. and full at 11.40 a.m.

This implies it takes 5 minutes for the liquid to double.

This is confirmed by the statement that before 5 minutes i.e. at 11.30 the bottle was only 1/4 full.

Hence true statements would be:

1) The bottle was 1/2 full at 11:35 a.m. and doubled again after 5 minutes.

2) The bottle was 1/4 full at 11:30 a.m. and doubled twice after 10 minutes. The bottle was 1/2 full at 11:35 a.m.

3) Exponential growth involves a constant multiplicative rate of change.

(The rate cannot be additional as we get 2 times every 5 minutes)

5 0

3 years ago

Read 2 more answers

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$