C. The value of the 7 in the hundredths place is 10 times the value of the 7 in the thousandths place.
$2.40 for binding
10 cents ($0.10) per page
cost=binding+pagescost
binding=2.40
pagescost=costperpage X number of pages=0.10*n if n=number of pages
so
cost=0.10n+2.40
or
C(x)=0.10n+2.40
the fixed cost is the binding
If these are terms of a geometric sequence, they have a common ratio. That is, ...
... (k -1)/(2(1 -k)) = (k +8)/(k -1)
... (k -1)² = 2(1 -k)(k +8) . . . . . multiply by the product of the denominators.
... k² -2k +1 = -2k² -14k +16 . . . eliminate parentheses
... 3k² +12k -15 = 0 . . . . . . . . put in standard form (subtract the right side)
... 3(k +5)(k -1) = 0 . . . . . . . . . factor
Possible values of k are ... -5, +1. The solution k=1 is extraneous, as it makes the first two terms 0 and the third term 8. (It doesn't work.)
The value of k is -5.
_____
The three terms are 12, -6, 3. The common ratio is -1/2.
Answer:
x = 4
-4x+21 => 5
Step-by-step explanation:
The given triangle in the picture is isosceles triangle which means that two sides of the triangle are equal.
So,
Now
Putting all the values in equation for perimeter
Putting x = 4 in side 1
Hence,
x = 4
-4x+21 => 5
Answer:
Below in bold.
Step-by-step explanation:
If the 3 points are on the circle with the given center then the distance from each point to the center will all be equal. This distance will be the radius of the circle.
Distance of P (2, 5) from the center (4, 2)
= √( (5-2)^2 + (2-4)^2)
= √(9 + 4)
= √13.
For Q (6, -1)this distance is:
= √(6-4)^2 + (-1-2)^2)
= √13.
For R (7, 4) his distance is:
= √(7-4)^2 + (4-2)^2)
= √13
These distances are the same so the points P, Q and R lie on the circle.
The midpoint of PQ
= (2+6)/2 , (5 - 1)/2
= (4, 2).
Which is the center so this verifies the second part.
