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ki77a [65]
3 years ago
14

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Mathematics
1 answer:
ki77a [65]3 years ago
8 0

️️️️️️️️️️️️️️️️️️️️️️️
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Determine and state an equation of the line parallel to the line 5x-4y=10 and passing through the point (4,12)
jonny [76]

Answer:

y=-2.5-5/4x and no it doesn't pass through (4,12)

Step-by-step explanation:

5x-4y=10

-5x     +5x

-4y=10+5x

/-4     /-4

y=-2.5-5/4x

3 0
3 years ago
The position function of a particle in rectilinear motion is given by s(t) = 2t^3 - 21t^2 + 60t + 3 for t ≥ 0. Find the position
avanturin [10]
Your first and second derivatives which are the velocity and the acceleration respectively are correct.
The particle has reverses direction when v(t) ≤  0 so at interval 2 < t < 5 sec.
You can substitute now t = 2 for s(2) and a(2) for first changing direction. Also t = 5 for s(5) and a(5) for second changing direction.
6 0
4 years ago
Which describes the slope of the given line?
ohaa [14]

Answer:

A. underlined

hccuv. huffy. kgigcgi igog if. lnhv

7 0
4 years ago
Read 2 more answers
50 points!! will mark brainliest to correct answer!!
Mamont248 [21]
I think your image for your regression line didn't show up, but judging from what I see, the answer should be:
<span>D) In step 4, Tameka mistakenly based the strength of the association on the slopes.

This is because the strength of the association is not determined by the slope of the regression line, but by how close the data points were to the actual line. The closer the points are to the actual regression line, the stronger the association.
</span>
4 0
3 years ago
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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