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3 years ago

FREE POINTS DONATION!!! 2 ANSWERS GETS IT!!! GET IT BEFORE SOME ELSE DOES!

Mathematics

2 answers:

True [87]3 years ago

4 0

Answer:

Step-by-step explanation:

Softa [21]3 years ago

3 0

Answer:

Thank you so much!!!!!!!!!!!

Step-by-step explanation:

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PLSSSSS HELPPPPPPPPP PLSSS

ra1l [238]

Answer:

Pretty sure it is A

Step-by-step explanation:

3 0

3 years ago

I NEED HELP ASAP PLS

maksim [4K]

Y=2/1x

Explanation:You start at -4 from the y and go to 6.So that becomes 10. Then you start at -2 from the x then go to 3 which mean you moved 5.So 10/5 simplified is is 2/1

8 0

2 years ago

Please someone help please pls

mario62 [17]

Answer:

h= $h= \frac{5R^{3} }{LU}$

Step-by-step explanation:

1.) You must isolate the h from the equation by dividing both sides by L (or multiply by its reciprocal, which in this case, the reciprocal on L is $\frac{1}{L}$ .....So it would look like:

$(\frac{1}{L} )*\frac{5R^{3} }{U} = \frac{L}{h}* (\frac{1}{h} )\\$

2.) Simplify the equation if you need to (in this case, its not needed)

$\frac{5R^3}{LU} =h$

That's your answer :)

3 0

3 years ago

A safety regulation states that the height hh of a guardrail should be 106 centimeters with an absolute deviation of no more tha

iren [92.7K]

Let us say that h is the height of the guardrail. Therefore the inequality equation that we can generate from this scenario is:

| h – 106 | = ± 7

There are two ways to solve this, either the equation is positive or negative.

When the equation is positive, therefore:

| h – 106 | = 7

h = 7 + 106 = 113 cm

When the equation is negative, therefore:

| h – 106 | = - 7

h = -7 + 106 = 99 cm

So the height must be 99 cm to 113 cm

6 0

3 years ago

Please help w this! Its a calculus question! look at the picture for the problem,

neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

$x^2+y^2=36\implies x=\sqrt{36-y^2}$

Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18$

Then the area of the entire blue region is 4 times this, a total of $\boxed{36\pi-72}$ .

Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18$

so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is

$\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18$

so the total area is, once again, $36\pi-72$ .

An even simpler way is to subtract the area of the square from the area of the circle.

$\pi6^2-(6\sqrt2)^2=36\pi-72$

6 0

3 years ago