How do you do this math

Mathematics

2 answers:

Basile [38]2 years ago

8 0

Answer: y = -3x - 6

Step-by-step explanation:

One way to write a linear equation is with slope-intercept form. Slope-intercept form is y = mx + b, where m is the slope, and b is the y-intercept.

Thus, the equation is y = -3x - 6

Hope it helps :) and let me know if you are confused anywhere.

professor190 [17]2 years ago

3 0

Answer:

y = -3x - 6

Step-by-step explanation:

Insert the information into the slope-intercept form equation:

y = mx + b

m = slope
b = y-intercept value

In this instance, the slope is -3 and the y-intercept value is -6. Therefore, the equation of the line is y = -3x - 6

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Write the equation of the circle with a radius 7 and center (13, 15). HELP ASAP!!!

ruslelena [56]

Answer:

Option b is the correct answer

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Step-by-step explanation:

$(x - 13)^2 + (y - 15)^2 = 7^2$

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2 years ago

The variable used to predict changes in the values of another value is called the response variable.

Sholpan [36]

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3 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .