Q 21. Simplify: (√5 + √2)
2
Q 22. Find the value of ρ(x) : ρ(x)= 5x2– 3x + 7 at x = 1
Q 23. Find the remainder when x3
+ 3x2
+ 3x + 1 is divided by x + 1
Q 24. Find the value of k, if x – 1 is factor of 4x2
+ 3x2
– 4x + k
Q 25. Draw the graph of x+ y =7
Q 26. Write fifth postulate.
Q 27. Prove that sum of all angles of triangle is 1800
Q 28. The taxi fare in a city is as follows. For the first kilometer, the fare is Rs. 8/- and for the
subsequent distance it is Rs. 5/- per km. Taking the distance covered as x km and total fare
as Rs. Y., write a linear equation for this information, and draw its graph.
Answers
1..... ;)
=>
Using (a+b)²= a²+2ab+b²
=>(√5)²+2(√5)(√2)+(√2)²
=>5+2+2√10
=>7+2√10
2..... ;)
=>
p(x)=5x ^2 −3x+7
Thus,
p(1)=5(1) ^2 −3(1)+7
p(1)=5−3+7=9
3....... ;)
=>
Apply remainder theorem
=>x + 1 =0
=> x = - 1
Replace x by – 1 we get
=>x3+3x2 + 3x + 1
=>(-1)3 + 3(-1)2 + 3(-1) + 1
=> -1 + 3 - 3 + 1
=> 0
Remainder is 0
4...... ;)
=>
x - 1 is a factor of 4x^3 + 3x^2 -4x +k
then x=1 is one root of 4x^3 + 3x^2 -4x +k
put x= 1 4x^3 +3x^2 -4x +k = 0
=> 4 (1)^3 +3 (1)^2-4 (1) +k =0
=> 4 + 3 - 4 + k = 0
=> k = -3
5...... ;)
=>
x+y=7
Putting x=0
0+y=7
y=7, so (0,7)
Putting y=0
x+0=7
x=7 so (7,0)
6........ ;)
=> Didn't understood.!! :(
7........ ;)
=>
In order to prove that the sum of angles of a triangle is 180, you must know the theorems of angles of a triangle.
We know that, alternate interior angles are of equal magnitude.
This'll help us get the answer. Let us assume a triangle ABC. Now we have to substitute the angles.
∠PAB + ∠BAC + ∠CAQ = 180°. where PQ line parallel to side BC touching vertex A.
∠PAB = ∠ABC & ∠CAQ = ∠ACB BECAUSE alternate interior angles are congruent..
Hence we get, ∠ABC + ∠BAC + ∠ACB = 180° [Proved]
8...... ;)
=>
Total fare = y
Total distance covered = x
Fair for the subsequent distance after 1st kilometre = Rs 5
Fair for 1st kilometre = Rs 8
A/q
y = 8 + 5(x-1)
⇒ y = 8 + 5x - 5
⇒ y = 5x + 3
R.E.A..
Step-by-step explanation:
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