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astraxan [27]
2 years ago
12

How many solutions are there if m < 0 and p > k ? p + m| x + n| = k

Mathematics
1 answer:
neonofarm [45]2 years ago
4 0

9514 1404 393

Answer:

  2

Step-by-step explanation:

The solutions are ...

  p +m|x +n| = k . . . . . given

  m|x +n| = k -p . . . . . . subtract p

  |x +n| = (k -p)/m = (p -k)/-m . . . . . . must have (p-k)/-m ≥ 0

  x = -n ± (p -k)/-m

For the given conditions, m < 0, p > k, we are guaranteed that (p-k)/-m > 0, so there are 2 solutions.

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Answer:

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Step-by-step explanation:

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a= 2, b= -5 and c= 1

x= (-b <u>+</u> √(b²-4ac))/2a

= (5 <u>+</u> √(25-8))/4

= (5 <u>+</u> √17)/4

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5 0
4 years ago
Read 2 more answers
Please don't take this down, it's just an innocent question
hichkok12 [17]

Answer:

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4 0
3 years ago
I don't get this can somone help me please ​
wolverine [178]

Answer:

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Step-by-step explanation:

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Hope this helps!

7 0
3 years ago
Can someone help and explain ? thank you !
AfilCa [17]
\sf Hello!

\sf y = 2x^{2} + 4x - 3

\sf Discriminant :

= \sf b^{2} - 4ac

= \sf (4)^{2} - 4.(2).(-3)

= \sf 16 - (-24)

= \sf 16 + 24 = \sf 40

\sf Since,
\sf D \:is\:Greater\: than\: 0.

\sf Roots\: are\: Real\: and\: Distinct

\sf Hence,

\sf y = 2x^{2} + 4x - 3\: have\: A.\: Two\: real\: roots

~ \sf iCarl
7 0
4 years ago
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