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stiv31 [10]
2 years ago
11

I have tried this for almost 20 mins , I might give up soon

Mathematics
1 answer:
seraphim [82]2 years ago
6 0

Answer:

x=1

y=3

z=-2

Step-by-step explanation:

Use substitution to solve two systems and then the third.

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Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer. How many imaginary r
Rainbow [258]

Answer:

<h3>The given polynomial of degree 4 has atleast one imaginary root</h3>

Step-by-step explanation:

Given that " Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer:

<h3>To find how many imaginary roots does the polynomial have :</h3>
  • Since the degree of given polynomial is 4
  • Therefore it must have four roots.
  • Already given that the given polynomial has 1 positive real root and 1 negative real root .
  • Every polynomial with degree greater than 1  has atleast one imaginary root.
<h3>Hence the given polynomial of degree 4 has atleast one imaginary root</h3><h3> </h3>

8 0
3 years ago
Whats the answer and why?
san4es73 [151]

A) -13

Bcz

x^2-13x+12=0

x^2-x-12x+12=0

x(x-1)-12(x-1)=0

(x-1)(x-12)=0

6 0
3 years ago
Please help me, thank you
Pavel [41]

Answer:

the answer is -3. Your welcome

4 0
2 years ago
DESPERATE NEED HELP PLEASEEE (08.01)
HACTEHA [7]
Answer:

<span>(−2, 3), because both lines pass through this point

----------------------------------
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5 0
3 years ago
EASY PLS HELP! 6= -2(7-c)
Vaselesa [24]
The answer is c = 10
3 0
2 years ago
Read 2 more answers
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