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Alexeev081 [22]
2 years ago
8

Answer this equation 3x+y=-3​

Mathematics
1 answer:
bearhunter [10]2 years ago
3 0

Answer:

3x = -y - 3

divide both sides by 3

x = -1/3y - 1

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a line that includes the point (3,4) has a slope of 2. what is the equation in the slope-intercept form?
Andreas93 [3]

Answer:

y=2x-2

Step-by-step explanation:

<em>the slope is y/x</em>

the slope is 2 or 2/1

<em>subtract 2 from the y value and 1 from the x value</em>

(3-1,4-2) = (2,2)

<em>keep doing this until you get a 0 in the x value</em>

(2-1,2-2) = (1,0)

<em>1 is your x-intercept</em>

(1-1,0-2) = (0,-2)

-2 is your y intercept.

So now you know your y-intercept and your slope so you can now write your equation.

<em>y=mx+b</em>

<em>m=slope, b=y-intercept</em>

m=2, b=-2

<em>substitute into the equation</em>

y=2x-2

8 0
3 years ago
What is the equation of the line that passes through the point (1,7) and is perpendicular to the graph of Y=-14X+11
vlada-n [284]
Y=1/14x might be the answer
4 0
3 years ago
How much will you have contributed to a Whole Life insurance policy that has a face value of $60,000 over a 40-year
boyakko [2]

Answer: The answer is $44,496. I got it right on the test. Sorry that I was late...

Step-by-step explanation:

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6 0
3 years ago
Need help with Calculus 1 inverse trig functions
lidiya [134]

Answer:

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}

Where u is a function of x as provided:

y=3tan^{-1}(x+\sqrt{1+x^2})

If we set

u=(x+\sqrt{1+x^2})

Then

\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}

\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}

Taking the derivative of y

y'=3[tan^{-1}(x+\sqrt{1+x^2})]'

Using the change of variables

\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}

\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}

Operating

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}

\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}

8 0
3 years ago
Joel needs 785 tickets to get the prize he wants at the arcade. He already has 284 tickets and can earn 35 tickets each time he
Cloud [144]

He needs to play 23 more times. 785 / 35 = 23

He will have extra points but it will be enough to get the prize.

4 0
2 years ago
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