A. y=180-15m
m is the number of months. 180 is the amount owed after the down payment was made.
B. X intercept (12,0)
Y intercept (0,180)
X intercept is the number of months until the tv is paid off. The y intercept is the amount owed after the down payment is made. How much they will be making payments on.
32. 5x + 15
Answer:
24
Step-by-step explanation:
simplify 8/10 into 2/5 then 2/5 of 30 is 24
1. find the area of the two triangles, including the shaded section. the formula for this is a=1/2bh
the height and base are both 24 inches. this is because you add the 10 from the side of the square to the 14 that is given
so:
a=1/2(24)(24)
a=(12)(24)
a=288 sq. inches
since there are two triangles, you would multiply the area by 2
a=2(288)
a=576 sq. inches
now, since you only need the unshaded section, you have to take away the shaded section, which is a square. to do this, you must calculate the area of the square and take it away from the area of both triangles.
a=576-(lw)
a=576- (10)(10)
a=576-100
a=476 sq. inches
that is your answer
Answer:
−3.632 > −3.792
|−3.632| < |−3.792|
Step-by-step explanation:
Given :
−3.632 −3.792 |−3.632| |−3.792|
First comparison :
−3.632 −3.792
Using the knowledge of number line :
- 3.792 occurs farther to the left of a number line than - 3.632, and numbers which occurs farther to the left are lesser. Therefore,
−3.632 > −3.792
2.)
|−3.632| |−3.792|
The presence of the absolute value symbol annuls the negative sign in the digit given. Hence, we can view both numbers as being positive values as ; 3.632 3.792
Numbers further to the right of a number line are greater and 3.792 occurs farther to the right than 3.632.
Hence,
|−3.632| < |−3.792|
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
