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3 years ago

HELP PLEASE 20 POINTS 1) What is the equation of the line in slope-intercept form?

Mathematics

2 answers:

Nezavi [6.7K]3 years ago

8 0

Answer:

y = (-5/4)x + 5

Step-by-step explanation:

Slope-intercept form is y = mx + b

m is the slope and b is the y-intercept

To find m, we can use the plotted points on the graph

(0,5)(4,0)

Now, do delta y/ delta x

(0-5)/(4-0)

slope = -5/4

Now, the y-intercept is the point where the graph crosses the y-axis

This point is where x is 0. Here, when x is 0, y is 5.

Therefore, the y-intercept is 5

Finally, replace what you've found into the equation

y = (-5/4)x + 5

Eddi Din [679]3 years ago

3 0

Answer:

y=-5/4x+5

Step-by-step explanation:

5/4 is the slope and 5 is the y-intercept

Hope this helped <3

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3 years ago

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Answer:

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Step-by-step explanation:

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Read 2 more answers

Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

slava [35]

$\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx$

Taking $x=7\tan\theta$ gives $\mathrm dx=7\sec^2\theta\,\mathrm d\theta$ , so that the integral becomes

$\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta$
$=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta$

When $\sec\theta>0$ , we have

$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta$
$=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta$

and from here we can substitute $u=\tan\theta$ to proceed from here.

Quick note: When we set $x=7\tan\theta$ , we are implicitly enforcing $-\dfrac\pi2$ just so that the substitution can be undone later via $\theta=\tan^{-1}\dfrac x7$ . But note that over this domain, we automatically guarantee that $\sec\theta>0$ , so the absolute value bars can be dropped immediately.

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3 years ago