If he answered 3 questions correctly, then the number of questions is got wrong is 7 and the number of questions that he did not attempt is 5.
<h3>What is the linear system?</h3>
A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.
In a class test of 15 questions, (+4) marks are given for every correct answer, (– 2) marks are given for every incorrect answer, and 0 marks are given for questions that are not attempted. Punit scored (– 2) marks.
If he answered 3 questions correctly, then the number of the question are got wrong and the number of the question that he did not attempt.
Let x be the correct question and y be the incorrect question, z be the number of questions that he did not attempt, and A be the number of the total score. Then we have
4x – 2z = A
x + y + z = 15
If A = -2 and x = 3, then we have
4 × 3 – 2z = -2
12 + 2 = 2z
z = 7
Then the value of y will be
3 + y + 7 = 15
y = 5
More about the linear system link is given below.
brainly.com/question/20379472
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Answer:
If a+b+c=1,
a
2
+
b
2
+
c
2
=
2
,
a
3
+
b
3
+
c
3
=
3
then find the value of
a
4
+
b
4
+
c
4
=
?
we know
2
(
a
b
+
b
c
+
c
a
)
=
(
a
+
b
+
c
)
2
−
(
a
2
+
b
2
+
c
2
)
⇒
2
(
a
b
+
b
c
+
c
a
)
=
1
2
−
2
=
−
1
⇒
a
b
+
b
c
+
c
a
=
−
1
2
given
a
3
+
b
3
+
c
3
=
3
⇒
a
3
+
b
3
+
c
3
−
3
a
b
c
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
c
a
)
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
(
a
b
+
b
c
+
c
a
)
+
3
a
b
c
=
3
⇒
(
1
×
(
2
−
(
−
1
2
)
+
3
a
b
c
)
)
=
3
⇒
(
2
+
1
2
)
+
3
a
b
c
=
3
⇒
3
a
b
c
=
3
−
5
2
=
1
2
⇒
a
b
c
=
1
6
Now
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
2
c
−
2
b
c
2
a
−
2
c
a
2
b
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
c
(
b
+
c
+
a
)
=
(
−
1
2
)
2
−
2
×
1
6
×
1
=
1
4
−
1
3
=
−
1
12
Now
a
4
+
b
4
+
c
4
=
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
2
2
−
2
×
(
−
1
12
)
=
4
+
1
6
=
4
1
6
Extension
a
5
+
b
5
+
c
5
=
(
a
3
+
b
3
+
c
3
)
(
a
2
+
b
2
+
c
2
)
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
c
2
)
]
=
3
⋅
2
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
]
Now
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
=
a
2
b
2
(
a
+
b
)
+
b
2
c
2
(
b
+
c
)
+
c
2
a
2
(
a
+
c
)
=
a
2
b
2
(
1
−
c
)
+
b
2
c
2
(
1
−
a
)
+
c
2
a
2
(
1
−
b
)
=
a
2
b
2
+
b
2
c
2
+
c
2
a
2
−
(
a
2
b
2
c
+
b
2
c
2
a
+
c
2
a
2
b
)
=
−
1
12
−
a
b
c
(
a
b
+
b
c
+
c
a
)
=
−
1
12
−
1
6
⋅
(
−
1
2
)
=
0
So
a
5
+
b
5
+
c
5
=
6
−
0
=
6
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
8 + 6 - 12 = 2
Answer:
4y+6y
Step-by-step explanation
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