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Tems11 [23]
3 years ago
10

Find a single expression that represents the distance from Adam's house to the corner if the distance between Adam's house and J

im's house is represented by the expression "23x − 2."

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
8 0
Jim's house to the corner is A. Adam's house to the corner is B. Adam's to Jim's is C.

(AxA)+(BxB)=(CxC). 

(15x-7)(15x-7) = 225x^2-210x+49 = (AxA)
(23x-2)(23x-2) = 529x^2-92x+4 = (CxC)

(CxC)-(AxA) = (BxB)

529x^2-92x+4-225x^2-210+49 = 304x^2-118x-45

so (BxB) = 304x^2-118x-45
now we would factor that expression but this one isn't factorable.

so your final answer for the distance from Adam's house to the corner is the square root of 304x^2-118x-45. 


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2 years ago
Given a point on a line and the equation of a parallel or perpendicular line, write the equation of the line in point-slope form
Arlecino [84]

Answer: y - 5 = 0(x - 1)

==================================================

Explanation:

Recall that point slope form in general is written as such

y - y1 = m(x - x1)

where,

m is the slope

(x1,y1) is the point the line goes through

The given equation y = 7 can be written as y = 0x+7. So we see that this line has a slope of m = 0

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which is the final answer

note: the equation in bold can be rearranged and simplified to get y = 5; however your teacher seems to want the answer in point-slope form, so we leave it as such.

6 0
3 years ago
5 less than a number is no more than 12
Naya [18.7K]
The answer should be 6 bro
4 0
3 years ago
SQRT is a parallelogram. If m∠QST = 72°, which of the following statements is true?
Mrrafil [7]
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Read 2 more answers
How many ways are there for six dogs and four cats to sit in a row in a strawberry field so that no two cats sit next to each ot
pickupchik [31]

Answer: 5,760

Step-by-step explanation:

\\It was given that each dog is distinct from other dogs and each cat is distinct from other cats, Also from the hint given, the First position is for the dogs. \\Let D represent the dogs and C represent the Cat , then we have

\\D C D C D C D C D D   or

\\D  D C D C D C D C D  or

\\D C DD C D C D C D or

\\D C D C D D C D C D or

\\D C D C D C DD C D

\\Each of the arrangements above  could be done in

\\2 x 4! x 4!  ( it is constant that D is starting , so I am only left with the arrangement of the remaining 5 D's  , out of the remaining 5 D's it is also constant that tow of them will be together and this could be done in 2 ways, so I have 4! left for the D's and 4! also for the C's )

\\= 2 x 24 x 24

\\ = 1 , 152

\\The total arrangement = 5 x 1 , 152

\\= 5,7 60 ways

3 0
3 years ago
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