Question

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Answer 1

Provided that:

• Final velocity = 80 kmph
• Initial velocity = 60 kmph
• Time taken = 5 seconds

To calculate:

• Acceleration

Solution:

• Acceleration = -1.112 mps sq.

Using concepts:

• Acceleration formula
• Formula to convert kmph-mps

Using formulas:

• a = (v-u)/t
• 1 kmph = 5/18 mps

Where, a denotes acceleration, v denotes final velocity, u denotes initial velocity and t denotes time taken.

Required solution:

~ Firstly let us convert kmph-mps!

Converting 60 kmph-mps!

→ 1 kmph = 5/18 mps

→ 60 kmph = 60 × 5/18 mps

→ 60 kmph = 300/18 mps

→ 60 kmph = 16.66 mps

• Henceforth, converted!

Converting 80 kmph-mps!

→ 1 kmph = 5/18 mps

→ 80 kmph = 80 × 5/18 mps

→ 80 kmph = 400/18 mps

→ 80 kmph = 22.22 mps

• Henceforth, converted!

~ Now let us calculate the acceleration!

→ a = (v-u)/t

→ a = (16.66-22.22)/5

→ a = -5.56/5

→ a = -1.112 mps sq.

→ Acceleration = -1.112 mps sq.

→ Retardation = -1.112 mps sq. $\:$

Answer 2

Answer:

P(x)=2x4−5x3+2x2−x+2andx2−3x+2=(x−1)(x−2)letx2−3x+2isthefactorofP(x)then,x2−3x+2=0⇒(x−1)(x−2)=0⇒x=1and2forx=1P(1)=0andforx=2P(2)=0∵P(1)=P(2)=0∴P(x)isdivisiblebyx2−3x+2Without actual division, prove that 2x4 – 5x3

+ 2x2 – x + 2 is divisible by x2 – 3x + 2

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Without actual division, prove that 2x4 – 5x3

Without actual division, prove that 2x4 – 5x3 + 2x2 – x + 2 is divisible by x2 – 3x + 2Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.

Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term]

Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term]= x(x-2)-1 (x-2)= (x-1)(x-2)

Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term]= x(x-2)-1 (x-2)= (x-1)(x-2)Hence, 0 of x2-3x+2 are land 2.

Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term]= x(x-2)-1 (x-2)= (x-1)(x-2)Hence, 0 of x2-3x+2 are land 2.We have to prove that, 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 i.e., to prove that p (1) =0 and p(2) =0

Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term]= x(x-2)-1 (x-2)= (x-1)(x-2)Hence, 0 of x2-3x+2 are land 2.We have to prove that, 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 i.e., to prove that p (1) =0 and p(2) =0Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 =2-5+2-1+2=6-6=0

Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term]= x(x-2)-1 (x-2)= (x-1)(x-2)Hence, 0 of x2-3x+2 are land 2.We have to prove that, 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 i.e., to prove that p (1) =0 and p(2) =0Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 =2-5+2-1+2=6-6=0and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0

Let p(x) = 2x4 – 5x3 + 2x2 – x+ 2 firstly, factorise x2-3x+2.Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term]= x(x-2)-1 (x-2)= (x-1)(x-2)Hence, 0 of x2-3x+2 are land 2.We have to prove that, 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2 i.e., to prove that p (1) =0 and p(2) =0Now, p(1) = 2(1)4 – 5(1)3 + 2(1)2 -1 + 2 =2-5+2-1+2=6-6=0and p(2) = 2(2)4 – 5(2)3 + 2(2)2 – 2 + 2 = 2x16-5x8+2x4+ 0 = 32 – 40 + 8 = 40 – 40 =0Hence, p(x) is divisible by x2-3x+2.