If the endpoints of a diameter are (6,3) and (2,1) the midpoint is the center of the circle so:
(x,y)=((6+2)/2, (3+1)/2)=(4,2)
Now we need to find the radius....the diameter is:
d^2=(6-2)^2+(3-1)^2
d^2=16+4
d^2=20 since d=2r, r=d/2, and r^2=d^2/4 so
r^2=5
The standard form of the circle is (x-h)^2+(y-k)^2=r^2 and we know:
(h,k)=(4,2) from earlier so:
(x-4)^2+(y-2)^2=5
Answer:
Let two consecutive multiples of 3 be x and (x+3)
A/q,
x * (x+3) = 648
➡ x² + 3x = 648
➡ x² + 3x -648 = 0
➡ x² + 27x - 24x -648 = 0
➡ x ( x + 27 ) -24 ( x +27)
➡ ( x - 24) ( x + 27)
➡ x = 24 and x = -27
so, we take x = 24.
Required multiples of 3
➡ x = 24
➡ x +3 = 24+3 = 27.
Odd = odd × odd
Prime factor
<span>16*24*60
(4*4)(3*8)(6*10)
(2*2*2*2)(3*2*2*2)(3*2*5*2)
multiple all the odd prime factors together
3*3*5 = 45
</span>
Answer:
Answer:
It can be A. (not B or C)
Step-by-step explanation:
It is A because x-6=0 can be simplified to x=6. Then, 2x + 6, you can divide the whole equation resulting in x+3=0, simplify this and you get x=-3. YES
It is not B because, while x+3=0 results in a zero of -3, 6x-1 can be simplified to be divided by 6. When we do this we get x-1/6=0, which is not equivalent to 6. NO
It is not C because, while x-6=0 results in a zero of 6, -3x can be simplified with the zero product property to get -3x=0 then dividing -3 by 0 giving you 0 which is not equivalent to one. NO
Step-by-step explanation:
Hello there! Your answer is approximately 14 students.
To solve this question, you want to start by finding 40% of 24. To do this, convert 40% into a decimal (divide by 100) and multiply by 24.
24 x 0.40 = 9.6.
But - you can't have 9.6 of a student. So, we round up to 10.
Note that the question asks how many students <em>still need to complete</em> their assignment, meaning we are looking for the ones that haven't finished. We found the amount for students that <em>have</em> finished, so subtract that from the total to see approximately how many still have to finish.
24 - 10 = 14
So, this means approximately 14 students still need to complete their assignment.
I hope I could help you, have a great rest of your day! :D
