Answer:
Equation of the circle (x-3)²+(y-5)²=(6.4)²
x² -6x +9 +y² -10y +25 = 40.96
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given endpoints of diameter P(−2, 1) and Q(8, 9)
Centre of circle = midpoint of diameter
Centre = 
Centre (h, k) = (3 , 5)
<u><em>Step(ii):-</em></u>
The distance of two end points
PQ = 
PQ = √164 = 12.8
Diameter d = 2r
radius r = d/2
Radius r = 6.4
<u><em>Final answer:-</em></u>
Equation of the circle
(x-h)²+(y-k)² = r²
(x-3)²+(y-5)²=(6.4)²
x² -6x +9 +y² -10y +25 = 40.96
x² -6x +y² -10y = 40.96-34
x² -6x +y² -10y -7= 0
Answer:
A. 5.4
Step-by-step explanation:
I think... I haven't done this sort of question in a while
Let's begin by listing out the information given to us:
There are four students: n = 4
Number of students to be selected: r = 2
To calculate the combination of 2 students to be chosen, we use:
Therefore, there 12 possible combinations from these
Example Problems<span>Find the value of "x": 5x = 25. Step 1 : The product is 25 and the given divisor is 5 . Given 5x = 25 . Divide by 5 on both the sides. 5 x 5 = 25 5 . ...<span>Find the value of "x" : 7x = 56; Step 1 : The product is 56 and the given divisor is 7. Given 7x = 56 . Divide by 7 on both the sides. 7 x 7 = 56 7 .</span></span>
Answer:
we have, 1953125=5⁹, so it cannot be a perfect square. If the last digit of a given number is 5, then the last three digits must be perfect squares, 025 or 225 or 625. Otherwise, that number cannot be a perfect square. And as 125 is not a perfect square, so no number ending with 125 can be a perfect square
