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blagie [28]
3 years ago
13

Question Help A television programmer is arranging the order that five movies will be seen between the hours of 6 P.M. and 4 A.M

. One of the movies has a G​ rating, and it is to be shown in the first time block. One of the movies is rated​ NC-17, and it is to be shown in the last of the time​ blocks, from 2 A.M.A.M. until 4 A.M. Given these​ restrictions, in how many ways can the five movies be arranged during the indicated time​ blocks?
Mathematics
1 answer:
Mamont248 [21]3 years ago
6 0
<span>During the indicated time block of 6 PM-4 AM, the movies should go as follows: Since the G rating movies needs to be shown during the first time block, it would be first at 6 pm until around 8 pm. The NC-17 movie would also need to be LAST and from 2 am to 4 am. That leaves 3 movies left to be shown from around 8 pm to 2 Am, which could be shown in 6 different ways. If we assigned EACH movie a title of A, B or C, you get these different showing possibilities- ABC, ACB, BAC, BCA, CAB, and CBA. Since the first and last movies never change, they would default to the first and last positions while the rest is filled in.</span>
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I don't understand this factorisation <br> a2+ 4a+3​
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Answer:

\boxed{\sf (a + 3)(a + 1)}

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3 years ago
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Answer:

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3 0
3 years ago
The polynomial x 3 + 5x 2 - ­57x -­189 expresses the volume, in cubic inches, of a shipping box, and the width is (x+3) in. If t
Ipatiy [6.2K]
V=x^3+5x^2-57x-189
Width: W=(x+3) in = 15 in →x+3=15
Solving for x:
x+3-3=15-3→x=12

With x=12 the Volume would be:
V=(12)^3+5(12)^2-57(12)-189
V=1,728+5(144)-684-189
V=1,728+720-684-189
V=1,575

V=W*D*H
Depth: D
Height: H
with H>D

V=1,575; W=15
Replacing in the equation above:
1,575=15*D*H
Dividing both sides by 15
1,575/15=(12*D*H)/15
105=D*H
3*5*7=D*H
D<H
If D=5→H=3*7→H=21
If D=7→H=3*5→H=15

Answer: Option <span>C. height: 21 in. depth: 5 in.
</span>
Please, see the attached file for another form to solve the problem

4 0
3 years ago
Read 2 more answers
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