Question

2. H(t)=√x²+2x+4;h(2)
3. k(x)= 3x²-1/2x+4;k (-3)
4. f(x)= 2x²+5x-9;f(5x-2)
5. g(p) = 4ˣ; x = 3 / 2​

Answer 1

$2. Given \: h(t) = \sqrt{ x^{2} + 2x + 4 }$

$\red{ Value \: of \: h(2) } \\= \sqrt{ 2^{2} + 2 \times 2 + 4 } \\= \sqrt{4 + 4 + 4 }\\= \sqrt{ 12} \\= 2\sqrt{3}$

Therefore .,

$\red{ Value \: of \: h(2) }\green { = 2\sqrt{3}}$

$3. Given \: k(x) =3x^{2} - \frac{1}{2} x + 4$

$\red{ Value \: of \: k(-3) }\\= 3\times (-3)^{2} - \frac{1}{2} \times (-3) + 4 \\= 27 + \frac{3}{2} + 4 \\= 31 + \frac{3}{2}\\= \frac{ 62 + 3}{2} \\= \frac{65}{2}$

Therefore.,

$\red{ Value \: of \: k(-3) }\green { = \frac{65}{2}}$

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Answer 2

2.Givenh(t)=

x

2

+2x+4

\begin{lgathered}\red{ Value \: of \: h(2) } \\= \sqrt{ 2^{2} + 2 \times 2 + 4 } \\= \sqrt{4 + 4 + 4 }\\= \sqrt{ 12} \\= 2\sqrt{3}\end{lgathered}

Valueofh(2)

=

2

2

+2×2+4

=

4+4+4

=

12

=2

3

Therefore .,

\red{ Value \: of \: h(2) }\green { = 2\sqrt{3}}Valueofh(2)=2

3

3. Given \: k(x) =3x^{2} - \frac{1}{2} x + 43.Givenk(x)=3x

2

−

2

1

x+4

\begin{lgathered}\red{ Value \: of \: k(-3) }\\= 3\times (-3)^{2} - \frac{1}{2} \times (-3) + 4 \\= 27 + \frac{3}{2} + 4 \\= 31 + \frac{3}{2}\\= \frac{ 62 + 3}{2} \\= \frac{65}{2}\end{lgathered}

Valueofk(−3)

=3×(−3)

2

−

2

1

×(−3)+4

=27+

2

3

+4

=31+

2

3

=

2

62+3

=

2

65

Therefore.,

\red{ Value \: of \: k(-3) }\green { = \frac{65}{2}}Valueofk(−3)=

2

65

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