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Anarel [89]
2 years ago
11

Evaluate the expression. I -3 I

Mathematics
2 answers:
Alex17521 [72]2 years ago
4 0

Answer:

the answer is 3

Step-by-step explanation:

sp2606 [1]2 years ago
3 0

Answer: 3

Step-by-step explanation:

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PLEASE HELP
borishaifa [10]
The answer should be the second one y=37
6 0
3 years ago
Hai! What Is 4 Squared???
Nitella [24]

Answer:

16

Step-by-step explanation:

x^{2} = x multiplied by x

so 4^{2} = 16

4 0
2 years ago
Read 2 more answers
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
Hi!
goblinko [34]
You can always add or subtract same value to/from both sides of equation. The same is true about multiplication and division.

In your case you have to subtract x from both sides of equation in order to get rid of it on the left side:
x+2y-x=8-x
2y=8-x

But it's easier to remember that if you want to move some member of equation to other side, you just have to change it's sign. Let's practice a bit using your equation (I have added "plus" signs where they are usually omitted to better understend what's going on):
+x+2y=+8

Move 2y to the right side:
+x=+8-2y

Move 8 to the left side:
+x+2y-8=0

Move both x and 2y to the right side:
0=8-x-2y
8 0
3 years ago
The slope of the line that passes through the points 5,10 and 2,12 is
V125BC [204]

slope is -2/3

y2-y1/x2-x1

so the first y and the second y and subtract those

thn subtract the second x frm the first x

then divide the two numbers

6 0
3 years ago
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