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Dilpreet Kaur 5 days, 20 hours ago

Pie chart of agriculture in punjab

Pashupati Yadav 1 week, 6 days ago

39

Pashupati Yadav 2 weeks, 3 days ago

16

Jiya Sood 2 weeks, 6 days ago

A-B={0,3} B-A={1,4} So (A-B)U(B-A)={0,1,3,4}

Saurabh Sharma 3 weeks, 1 day ago

Ans- {0, 1,3,4}

Jiya Sood 2 weeks, 6 days ago

Let tye square root of 1+i =root over 1+i Root over 1+i=x+iy By Squaring both sides We get , 1+i=(x+iy)² 1+i=x²+y²i²+2xyi As we know that i²=-1 So,, 1+i=x²-y²+2xyi Now compare real part and imaginary part on both sides... 1=x²-y²----1 equation 1=2xy------2equatiin We know a identwe says that (x²+y²)²=(x²-y²)²+(2xy)² Now put the values in this identity On adding and subtracting the above equations , we get ∴x2=22​+1​ and y2=22​−1​. ∴x=±22​+1​​andy=±22​−1​​ So, by equation (1), 1−i​=22​+1​​−i22​−1​​ Ans: 1

Aryan Choudhary 4 weeks ago

Bjkkikgg

Mohit Kumar 1 week, 6 days ago

2

Prksh Sonali 1 month, 3 weeks ago

The focus of parabola is F ( 2,0 ) and it's directrix is the line x =-3

Preeti Dabral 2 months, 2 weeks ago

Solution is given below:

{tex}\begin{aligned} & n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ & n(B)=42-20+4=26 \\ & n(A-B)=n(A)-n(A \cap B)=20-4=16 \\ & n(B-A)=n(B)-n(A \cap B)=26-4=22 \end{aligned}{/tex}

32

Test Account 1 month, 1 week ago

32

Sharan M 2 months, 2 weeks ago

32

Md Mazid Mahboob 2 months, 2 weeks ago

K

Sharan M 2 months, 2 weeks ago

i

Preeti Dabral 2 months, 2 weeks ago

We are given that, p = 4 and ω = 150
Now, {tex}\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}{/tex}
and {tex}\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}{/tex}
The equation of the line is x cos ω + y sin ω = p
{tex}x \cos 15^{\circ}+y \sin 15^{\circ}{/tex}
or {tex}\frac{\sqrt{3}+1}{2 \sqrt{2}} x+\frac{\sqrt{3}-1}{2 \sqrt{2}} y=4{/tex}
or {tex}(\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}{/tex}
This is the required equation.

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