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3 years ago

Place labels on the coordinate plane below.

Mathematics

2 answers:

Nuetrik [128]3 years ago

8 0

Answer:

the x-axis is the middle horizontal line

the y-axis in the middle vertical line

the origin is (0,0) or where the x and y-axis cross

in the upper left quadrant the x-axis is negative and the y-axis is positive.

in the upper right quadrant the x-axis is positive and so is the y.

in the lower left quadrant both the x and y-axis is negative.

in the lower right quadrant the x-axis is positive and the y-axis is negative.

sergejj [24]3 years ago

3 0

Answer:

Step-by-step explanation:

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3 years ago

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

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Sliva [168]

Answer:

Step-by-step explanation:

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2 years ago

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3 years ago

One factor of f (x ) = 5 x cubed + 5 x squared minus 170 x + 280 is (x + 7). What are all the roots of the function? Use the Rem

PilotLPTM [1.2K]

The roots of the function f(x) = 5x³ + 5x² - 170x + 280 are given by x = 4, x = 2 or x = -7

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that x + 7 is the root of 5x³ + 5x² - 170x + 280, hence:

(5x³ + 5x² - 170x + 280) / (x + 7) = 5x² - 30x - 40

Hence:

(5x³ + 5x² - 170x + 280) = (x + 7)(5x² - 30x - 40)

= 5(x² - 6x - 8)(x + 7)

= 5(x - 4)(x - 2)(x + 7)

The roots are at:

0 = 5(x - 4)(x - 2)(x + 7)

x = 4, x = 2 or x = -7

The roots of 5x³ + 5x² - 170x + 280 are x = 4, x = 2 or x = -7

Find out more on equation at: brainly.com/question/1214333

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2 years ago