Jacob currently grows and sells 10 butternuts each week to a spaza shop. Each butternut costs R4 to grow and is sold for R7. Jacob notices that the spaza shop does not yet sell cauliflowers. Cauliflowers cost R6 each to grow. Jacob visited the spaza shop and discussed the possibility of selling cauliflowers with the owner. He is open to the idea and is prepared to buy one cauliflower a week for R12,50. He could buy more cauliflowers, but is then prepared to pay less for each of the cauliflowers as he would not be able to sell them all at such a high price. After thinking about it, he is prepared to buy the following quantities of cauliflowers per week at the following prices:

Prepared to buy 1 2 3 4 5 6 7 8 9 10
Price per cauliflower bought R12.50 R12 R11.50 R11 R10.50 R10 R9.50 R9 R8.50 R8

Jacob can only grow and sell 10 vegetables in a week. What is the optimum number of cauliflower and butternut he should sell to maximize his profit?

d

step-by-step explanation:

c. side angle side. u shares the same sides

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

​12

​

​11

​​   to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

​c

​

​b

​​ =

​c

​

​ac+b

​​

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

​3

​

​1

​​ +

​6

​​5

​​ x=

​12

​5×12+11

​​ 2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

​3

​​1

​​ +

​6

​5

​​ x=

​12

​​60+11

​​3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

​3

​​1

​​ +

​6

​​5

​​ x=

​12

​​71

​​4 simplify \frac{5}{6}x

​6

​​5

​​ x to \frac{5x}{6}

​6

​5x

​​ 3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

​3

​

​1

​​ +

​6

​5x

​​ =

​12

​​71

​​ 5 convert 1\frac{1}{3}1

​3

​​1

​​   to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

​c

​​b

​​ =

​c

​​ac+b

​​3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​1×3+1

​​ +

​6

​​5x

​​ =

​12

​71

​​ 6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​3+1

​​ +

​6

​​5x

​​ =

​12

​71

​​ 7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​4

​​ +

​6

​5x

​​ =

​12

​71

​​8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

​3

​​4

​​ +

​6

​​5x

​​   to \frac{13.75}{3}+\frac{4x}{3}

​3

​​13.75

​​ +

​3

​4x

​​ \frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

​3

​​13.75

​​ +

​3

​​4x

​​ =

​12

​71

​​ 9 simplify \frac{13.75}{3}

​3

​​13.75

​​   to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

​3

​​4x

​​ =

​12

​71

​​ 10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

​3

​

​4x

​​ =

​12

​

​71

​​ −4.583333

11 simplify \frac{71}{12}-4.583333

​12

​

​71

​​ −4.583333 to \frac{4}{3}

​3

​4

​\frac{4x}{3}=\frac{4}{3}

​3

​​4x

​​ =

​3

​4

​​ 12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

​3

​4

​​ ×3

13 simplify \frac{4}{3}\times 3

​3

​​4

​​ ×3 to \frac{12}{3}

​3

​​12

​​ 4x=\frac{12}{3}4x=

​3

​​12

​​ 14 simplify \frac{12}{3}

​3

​​12

​​   to 44

4x=44x=4

15 divide both sides by 44

x=1

​​

x-150=475   x=625

step-by-step explanation:

if the bike after the sale ($150) is $475 then just simply add 150 to 475.