Mike is buying new uniforms for work. He buys 2 uniforms, each of which include pair of pants for $34 and a shirt.His total was $106,without taxWhat is the equation for this situation?
Answers
2(p + s) = 106
2(34 + s) = 106
68 + 2s = 106
2s = 38
s = 19
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Mr. and Mrs. Zeller close on a 20 year home loan for $150,000. The monthly payment with no points is $1,160, but if they buy a point it is $1,150. What might you infer if Mr.and Mrs. Zeller choose not to buy a point?
There are 28 pens in a full box. how many pens are in 35 full boxes
Y = 3
5x^-6y^-2/y
Answers
Your answer is 5/27.
Answers
Answer:
I think B
Step-by-step explanation:
Answer:
A.) 0.32
Step-by-step explanation:
is 4 but h = 2 which is 1/2 of the answer
so is 6.324 and 0.32 is the closet answer
Answers
Rearrange the terms
x with x and numerals with numerals
2x - x + 3 + 5
x + 8
The simple form of given expression 2x + 3 − x + 5 is equal to (x+8).
Given that:
The expression: 2x + 3 - x + 5
To simplify the given expression, combine like terms. The terms with x can be combined, as well as the constant terms.
The simplified form is:
2x - x + 3 + 5
Combining the x terms gives as:
(2x - x) + 3 + 5
Simplifying further the above expression as:
x + 3 + 5
Adding the constant terms:
x + 8
Hence, the simplified form of 2x + 3 - x + 5 is x + 8.
Learn more about Expression here:
#SPJ6
Answers
y-weigh of Julie
z-weigh of Mike
x=2y
z=3y
x+y+z=210lbs
2y+y+3y=210
6y=210 /:6
y=35 lbs
x=2y=2*35=70 lbs
z=3y=3*35=105 lbs
Answers
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
Answers
Therefore, the annual depreciation is $1,199.0
Answer:
$1,199 per year.
Step-by-step explanation:
Tina Cole is a certified financial adviser.
She purchased equipment for her office costs = $8,843
After 7 years of use the trade-in value of the equipment is estimated = $450
Total depreciation in 7 years = 8,843 - 450 = $8,393
We have to calculate the annual depreciation using the straight-line method.
So we divide $8,393 by 7
$8,393 ÷ 7 = $1,199
The annual depreciation of the equipment is $1,199 per year.