The answer is x = 6
hope this helps
Answer:
(x, y) = (18, 5)
Step-by-step explanation:
Assuming the three lines meet at a single point at lower left (the figure is sloppily drawn), the angle (3x)°+49° is a corresponding angle to (7x-23)°. That means they have the same measure:
3x +49 = 7x -23
72 = 4x . . . . . . . . . add 23-3x
18 = x . . . . . . . . . . . divide by 4
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Angles (3x)° and (11y-1)° are "corresponding" angles, so are congruent.
3x = 11y -1
3(18) +1 = 11y . . . . add 1, fill in the value of x
55/11 = y = 5 . . . . divide by 11
The values of x and y are 18 and 5, respectively.
Answer:
The percent error of Heather's calculation is <u>8%</u>.
Step-by-step explanation:
Given:
Heather measures the temperature of her coffee to be 133.4 degrees fahrenheit. It is actually 145 degrees fahrenheit.
Now, to find the percent error of Heather's calculation.
The temperature of coffee Heather measures = 133.4° F.
Coffee's actual temperature = 145° F.
So, to get the measurement in error we subtract the temperature of coffee Heather measures from coffee's actual temperature:
Now, to get the percent error:
Therefore, the percent error of Heather's calculation is 8%.
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 !
