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3 years ago

What is the slope of the line 3x- 9y=4

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

6 0

Answer:

C) 1/3

Step-by-step explanation:

On Edge2020

vlabodo [156]3 years ago

4 0

Answer:

1/3 is your answer

Step-by-step explanation:

Note that in the slope intercept form, the equation looks like: y = mx + b

Isolate the y. First, subtract 3x from both sides

3x (-3x) - 9y = (-3x) + 4

-9y = -3x + 4

Next, to fully isolate the y, divide - 9 from both siedes

(-9y)/(-9) = (-3x + 4)/(-9)

y = (-3x + 4)/-9

Simplify

y = (1/3)x - 4/9

You're slope of the line (m in y = mx + b) is 1/3

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The population of rabbits on an island is growing exponentially. In the year 1995, the population of rabbits was 1000, and by 19

irakobra [83]

Answer:

3,240

Step-by-step explanation:

The computation of the population of rabbits in the year 2003 is shown below:

Given that

In the year 1995, the population of the rabbits was 1000

And, in 1999 the population of the rabbits grown to 1,800

So there is an increase of

= (1800 - 1000) ÷ 1000

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2 years ago

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Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

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lara [203]

Answer:

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Step-by-step explanation:

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2 years ago

I dont know how to solve this does anybody?

postnew [5]

Answer:

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Step-by-step explanation:

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