Answer:
k(x) = (3g + 5h)(x) ⇒ (1)
k(x) = (5h - 3g)(x) ⇒ (3)
k(x) = (h - g)(x) ⇒ (2)
k(x) = (g + h)(x) ⇒ (4)
k(x) = (5g + 3h)(x) ⇒ (5)
k(x) = (3h - 5g)(x) ⇒ (6)
Step-by-step explanation:
* To solve this problem we will substitute h(x) and g(x) in k(x) in the
right column to find the corresponding function formula in the
left column
∵ h(x) = 5 - 3x
∵ g(x) = -3^x + 5
- Lets start with the right column
# k(x) = (3g + 5h)(x)
∵ g(x) = -3^x + 5
∵ 3g(x) = 3[-3^x + 5] = [3 × -3^x + 3 × 5]
- Lets simplify 3 × -3^x
take the negative out -(3 × 3^x), and use the rule a^n × a^m = a^(n+m)
∴ -3(3 × 3^x) = -(3^x+1)
∴ 3g(x) = -3^x+1 + 15
∵ h(x) = 5 - 3x
∵ 5h(x) = 5[5 - 3x] = [5 × 5 - 5 × 3x] = 25 - 15x
- Now substitute 3g(x) and 5h(x) in k(x)
∵ k(x) = (3g + 5h)(x)
∴ k(x) = -3^x+1 + 15 + 25 - 15x ⇒ simplify
∴ k(x) = 40 - 3^x+1 - 15x
∴ k(x) = 40 - 3^x+1 - 15x ⇒ k(x) = (3g + 5h)(x)
* k(x) = (3g + 5h)(x) ⇒ (1)
# k(x) = (5h - 3g)(x)
∵ 5h(x) = 25 - 15x
∵ 3g(x) = -3^x+1 + 15
∵ k(x) = (5h - 3g)(x)
∴ k(x) = 25 - 15x - (-3^x+1 + 15) = 25 -15x + 3^x+1 - 15 ⇒ simplify
∴ k(x) = 10 + 3^x+1 - 15x
∴ k(x) = 10 + 3^x+1 - 15x ⇒ k(x) = (5h - 3g)(x)
* k(x) = (5h - 3g)(x) ⇒ (3)
# k(x) = (h - g)(x)
∵ h(x) = 5 - 3x
∵ g(x) = -3^x + 5
∵ k(x) = (h - g)(x)
∴ k(x) = 5 - 3x - (-3^x + 5) = 5 - 3x + 3^x - 5 ⇒ simplify
∴ k(x) = 3^x - 3x
∴ k(x)= 3^x - 3x ⇒ k(x) = (h - g)(x)
* k(x) = (h - g)(x) ⇒ (2)
# k(x) = (g + h)(x)
∵ h(x) = 5 - 3x
∵ g(x) = -3^x + 5
∵ k(x) = (g + h)(x)
∴ k(x) = -3^x + 5 + 5 - 3x ⇒ simplify
∴ k(x) = 10 - 3^x - 3x
∴ k(x)= 10 - 3^x - 3x ⇒ k(x) = (g + h)(x)
* k(x) = (g + h)(x) ⇒ (4)
# k(x) = (5g + 3h)(x)
∵ g(x) = -3^x + 5
∵ 5g(x) = 5[-3^x + 5] = [5 × -3^x + 5 × 5] = 5(-3^x) + 25
∴ 5g(x) = -5(3^x) + 25
∵ h(x) = 5 - 3x
∵ 3h(x) = 3[5 - 3x] = [3 × 5 - 3 × 3x] = 15 - 9x
- Now substitute 5g(x) and 3h(x) in k(x)
∵ k(x) = (5g + 3h)(x)
∴ k(x) = -5(3^x) + 25 + 15 - 9x ⇒ simplify
∴ k(x) = 40 - 5(3^x) - 9x
∴ k(x) = 40 - 5(3^x) - 9x ⇒ k(x) = (5g + 3h)(x)
* k(x) = (5g + 3h)(x) ⇒ (5)
# k(x) = (3h - 5g)(x)
∵ 3h(x) = 15 - 9x
∵ 5g(x) = -5(3^x) + 25
∵ k(x) = (3h - 5g)(x)
∴ k(x) = 15 - 9x - [-5(3^x) + 25] = 15 - 9x + 5(3^x) - 25 ⇒ simplify
∴ k(x) = 5(3^x) - 9x - 10
∴ k(x) = 5(3^x) - 9x - 10 ⇒ k(x) = (3h - 5g)(x)
* k(x) = (3h - 5g)(x) ⇒ (6)
![\sqrt[4][a^5 b^3]](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%5Ba%5E5%20b%5E3%5D)
) is the exponent.![\sqrt[][text]) symbol.\\a fourth root is a factor that was multiplied four times. ex: 2*2*2*2=16. the fourth root of 16 is two. \\when you do square roots, you take the number representing the factor out of the radical. anything else that is not a square root stays under. ex: \\([tex]\sqrt[9][text])\\([tex]\sqrt[3*3*3][text]) there is one set of two 3's so\\([tex]3\sqrt[3][text])\\with a fourth root, you look for groups of four. the symbol of that group is placed outside the radical. anything else stays in.\\ ([tex]\sqrt[4][a^5 b^3]](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%5Btext%5D%29%20symbol.%5C%5Ca%20fourth%20root%20is%20a%20factor%20that%20was%20multiplied%20four%20times.%20ex%3A%202%2A2%2A2%2A2%3D16.%20the%20fourth%20root%20of%2016%20is%20two.%20%5C%5Cwhen%20you%20do%20square%20roots%2C%20you%20take%20the%20number%20representing%20the%20factor%20out%20of%20the%20radical.%20anything%20else%20that%20is%20not%20a%20square%20root%20stays%20under.%20ex%3A%20%5C%5C%28%5Btex%5D%5Csqrt%5B9%5D%5Btext%5D%29%5C%5C%28%5Btex%5D%5Csqrt%5B3%2A3%2A3%5D%5Btext%5D%29%20there%20is%20one%20set%20of%20two%203%27s%20so%5C%5C%28%5Btex%5D3%5Csqrt%5B3%5D%5Btext%5D%29%5C%5Cwith%20a%20fourth%20root%2C%20you%20look%20for%20groups%20of%20four.%20the%20symbol%20of%20that%20group%20is%20placed%20outside%20the%20radical.%20anything%20else%20stays%20in.%5C%5C%20%28%5Btex%5D%5Csqrt%5B4%5D%5Ba%5E5%20b%5E3%5D)
)![\sqrt[4][a*a*a*a*a*b*b*b]](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%5Ba%2Aa%2Aa%2Aa%2Aa%2Ab%2Ab%2Ab%5D)
) there is one group of 4 a's so![a\sqrt[4][a* b^3]](https://tex.z-dn.net/?f=a%5Csqrt%5B4%5D%5Ba%2A%20b%5E3%5D)
)