Answer:
yes
Step-by-step explanation:
There are no repeating numbers in the domain
The equation of a circle centred at point (m,n) and radius r is given by
<span>(x-m)² + (y-n)² = r²
</span>-------------------------------------------------------------
Centre = (4,3)
radius = 5
Equation:
(x - 4)² + (y - 3)² = 5²
⇒ x² - 8x + 16 + y² - 6y + 9 = 25
⇒ x² + y² - 8x - 6y + 25 = 25
⇒ x² + y² - 8x - 6y = 0
The equation of the circle is x² + y² - 8x - 6y = 0
Hope it helps!
1a) Possible rational roots will be of the form
±{divisor of 10}/{divisor of 4}
Divisors of 10 are {1, 2, 5, 10}
Divisors of 4 are {1, 2, 4}
Then possible rational roots are
{±1/4, ±1/2, ±1, ±5/4, ±2, ±5/2, ±5, ±10}
1b) Your answer is correct.
2) One additional root will be the conjuate of the given complex root.
5 -3i
3) If one root is 5 -√7, another will be 5 +√7. Then your polynomial is
P(x) = (x -(5 -√7))*(x -(5 +√7)) = (x -5)^2 -(√7)^2
P(x) = x^2 -10x +18
answer
10
step-by-step explanation
the equation given is sin(x) = cos(y) with x = 2k + 3 and y = 6k + 7
substitute in 2k+ 3 for x in sin(x) and substitute in 6k + 7 for y in cos(y)
sin(x) = cos(y)
sin(2k + 3) = cos(6k + 7)
we know that sin(x) = cos(90 -x)
sin(2k + 3)
= cos(90 - (2k + 3) )
= cos(90 - 2k - 3)
= cos(87 - 2k)
substitute this into the equation sin(2k + 3) = cos(6k + 7)
sin(2k + 3) = cos(6k + 7)
cos(87 - 2k) = cos(6k + 7)
87 - 2k = 6k +7
80 = 8k
k = 10
Answer:
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