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FromTheMoon [43]
3 years ago
12

Hi, prove that (AXB).(CXD)+(BXC).(AXD)+(CXA).(BXD)=0

Mathematics
1 answer:
slava [35]3 years ago
5 0
No.i will not prove that.
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Please help in math!!
LenaWriter [7]

Answer: C

Step-by-step explanation:

Less than or more than without "or equal" means that the dot is not shaded, otherwise it is.

So the answer is C

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2 years ago
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Match the circle equations in general form with their corresponding equations in standard form. Not all will be used. 
Xelga [282]
<span>The standard form of the equation of a circumference is given by the following expression:

</span>(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius
<span>
On the other hand, the general form is given as follows:

</span>x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}<span>

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

</span>\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46

Equations in General Form:

\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0

\bold{6)} \ x^{2}+y^{2}+2x-12y

So let's match each equation:

\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4

Then, its general form is:

x^{2}+y^{2}-12x-8y-4=0

<em><u>First. a) matches 5) </u></em>

\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20

Then, its general form is:

x^{2}+y^{2}-4x+12y-20=0

<em><u>Second. b) matches 1) </u></em>

\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5

Then, its general form is:

x^{2}+y^{2}+4x+6y-5=0

<em><u>Third. c) matches 3)</u></em>

\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9

Then, its general form is: x^{2}+y^{2}+2x-12y-9=0

<em><u>Fourth. d) matches 6)</u></em>
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Answer:

90:180, 135:270, 180:360

   

Step-by-step explanation:

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If you want the intersect of M and N, then:

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Step-by-step explanation:

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