Answer:
see the explanation
Step-by-step explanation:
we know that
The circumference of a circle is equal to

where
D is the diameter
In this problem we have


substitute and solve for D



we know that
The length of the row of quarters is equal to multiply the diameter of the quarter by the number of quarters in the row
so
For 4 quarters in the row, the length is equal to --> 
For 6 quarters in the row, the length is equal to --> 
For 12 quarters in the row, the length is equal to -> 
Answer:
Step-by-step explanation:
r-13,725 <_ 1650
r-<_ 15,375
r-13725>_-1650
r >_12,125
Tuition could be between $12,125 and $15,375
PLS MARK ME AS BRAINLIEST
Answer:
Vertical A @ x=3 and x=1
Horizontal A nowhere since degree on top is higher than degree on bottom
Slant A @ y=x-1
Step-by-step explanation:
I'm going to look for vertical first:
I'm going to factor the bottom first: (x-3)(x-1)
So we have possible vertical asymptotes at x=3 and at x=1
To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)
3^3-5(3)^2+4(3)-25=-31 so it isn't a factor of the top so you have a vertical asymptote at x=3
Let's check x=1
1^3-5(1)^2+4(1)-25=-25 so we have a vertical asymptote at x=1 also
There is no horizontal asymptote because degree of top is bigger than degree of bottom
There is a slant asympote because the degree of top is one more than degree of bottom (We can find this by doing long division)
x -1
--------------------------------------------------
x^2-4x+3 | x^3-5x^2+4x-25
- ( x^3-4x^2+3x)
--------------------------------
-x^2 +x -25
- (-x^2+4x-3)
---------------------
-3x-22
So the slant asymptote is to x-1
Answer:
C
Step-by-step explanation:
Area of rectangle A is 2 and area of B is 18. 18 divided by 2 is 9. Therefore rectangle B is 9 times the area of rectangle A
Answer:

Step-by-step explanation:
Remember these rules for addition, substraction, multiplication or division:
Addition
+ + = +
- - = -
Big + number - = +
Big -number + = -
Substraction
change the sign!
And follow the rules for addition
Multiplication or division
+ + = +
- - = +
+ - = -
Now for the exercise

<u>- (
)</u>
<u />
Change the signal

<u>(
)</u>
