Uma jarra tem a forma de um paralelipipedo retangulo com20 cm de altura e 120cm 2 de área de base . A jarra tem água até sete sete sobre dez da sua altura. Quantos centilitros de água são nessecarios para encher a jarra ?

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above

your answer is 228

step-by-step explanation:

add the top and bottom bases together, then divide them by 2, then multiply by the height.

step-by-step explanation:

the shape,the size and the distance

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

​​