Answer:
We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.
We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :
1 - We first group terms with the same variable :
(x²+8x) + (y²+22y) + 37 = 0
2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)
(x²+8x) + (y²+22y) = - 37
3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.
x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²
y²+22y = (y²+22y+11²)-11² = (y+11)²-11²
4 - We plug the new values inside our equation :
(x+4)² - 4² + (y+22)² - 11² = -37
(x+4)² + (y+22)² = -37+4²+11²
(x+4)²+(y+22)² = 100
5 - We re-write in standard form :
(x-(-4)²)² + (y - (-22))² = 10²
And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)
Answer:
0.5
Step-by-step explanation:
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Given data
<span>sin (x+pi/2)=cos x
</span>now using sin law
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
now using above values
sin(pi/2+x)=sin(pi/2)cos(x)+cos(pi/2)sin(x)
as we know that
sin(pi/2)=1
cos(pi/2)=0
now putting these values
sin(pi/2+x)=1*cosx+0*1
sin(pi/2+x)=cosx
hence proved that
<span>sin (x+pi/2)=cos x</span>
Answer:
The answer is 8
- 3 + 2 - =<u> 4 </u> + <u>2 </u> .
Step-by-step explanation:
Given:
8x - 3x + 2 -x = __ x + __
Now, to write a number in each blank to complete each equation so that it has the requested number of solutions.
So, to solve the equation:
So, we add the variables:
And, then subtract it:
Thus, it remains:
.
So the complete equation is:
8 - 3 + 2 - =<u> 4 </u> + <u>2 </u>
Therefore, the answer is 8 - 3 + 2 - =<u> 4 </u> + <u>2 </u> .
<u />
Use the depreciation formula.
Where 'p' is the principal value, 'r' is the rate it depreciates, and 'n' is the time. Just plug in what we know:
Simplify by subtracting:
Simplify the exponent:
Multiply:
