Allison practices her violin for at least 12 hours per week she practices for three fourths of an hour each session if Allison has already practiced 3 hours this week how many more sessions remains for her to meet or exceed her weekly practice goals.
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Related Questions
PLEASE HELP WITH THE CORRECT ANSWERRRR PLEASEWhat value of b makes the trinomial below a perfect square?x^2-bx + 100A. 20B. 10C. 5D. 50
They Supreme Court chooses to hear cases that (blank)
There are 25 students in Mrs. Venetozzi’s class at the beginning of the school year, and the average number of siblings for each student is 3. A new student with 8 siblings joins the class in November.The average number of siblings that each student has is
Using the fact that cos(52) = 0.6157, find the measure of angle B if sin(B) = 0.6157.
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There are a couple of things you didn't mention.
In order to come up with an answer to the question,
I have to assume reasonable numbers for them, and
then answer the question that I have invented.
Assumption #1: Each day that you work, you work for 8 hours.
Assumption #2: You work for three days each week.
If those assumptions are true, then you earn
($7.25 / hour) x (8 hours/day) x (3 days/week) x (2 weeks) =
($7.25 x 8 x 3 x 2) = $348 before taxes and other deductions.
5 yd
5 yd
a
500 yd2
О Б
100 yd2
Oc
50 yd?
Od
200 yd?
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Answer:
a- 500yd^3
Step-by-step explanation:
volume=lwh
=5x5x20
=500
brainliest appreciated!!
(x + 4) and (x – 4)
(x + 3) and (x – 4)
(x + 4) and (x – 1)
(x + 3) and (x – 1)
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The quadratic trinomial can be factor using the following formula:
.
1. For the quadratic trinomial find the roots:
.
2. The factoring form of trinomial is
.
3. The factors are x+4 and x-1
Answer: correct choice is C.
factors of x² are x * x
factors of -4 that will result to a +3
4 * -1 = -4
4 - 1 = 3
(x + 4) and (x - 1)
x(x-1) + 4(x-1)
x² - x +4x - 4
x² + 3x - 4
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Answer:
(1,1)
Step-by-step explanation:
use the midpoint formula to find the midpoint of the line segment
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Answer:
88
Step-by-step explanation:
Write the expression for the sum in the relation you want.
Sn = u1(r^n -1)/(r -1) = 2.1(1.06^n -1)/(1.06 -1)
Sn = (2.1/0.06)(1.06^n -1) = 35(1.06^n -1)
The relation we want is ...
Sn > 5543
35(1.06^n -1) > 5543 . . . . substitute for Sn
1.06^n -1 > 5543/35 . . . . divide by 35
1.06^n > 5578/35 . . . . . . add 1
n·log(1.06) > log(5578/35) . . . take the log
n > 87.03 . . . . . . . . . . . . . . divide by the coefficient of n
The least value of n such that Sn > 5543 is 88.
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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 !
Answer:
Step-by-step explanation:
362,880