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seropon [69]
3 years ago
13

A group of people attempt to bake the largest vegan cake the used 17 kilograms of cocoa powder which is about one-tenth the amou

nt of dates. How many kilograms of cocoa powder and dates do they use?
Mathematics
1 answer:
Fiesta28 [93]3 years ago
7 0

Answer: 187 kilogram

Step-by-step explanation:

Let the kilogram of dates be represented by x.

Since they used 17 kilograms of cocoa powder which is about one-tenth the amount of dates. The dates will be:

= 1/10 × x = 17

0.1x = 17

x = 17/0.1

x = 170 kilogram

Date = 170 kilogram.

Cocoa powder = 17 kilogram

The total kilogram of date and cocoa powder used will be:

= 170 + 17

= 187 kilogram

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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)

4 0
3 years ago
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