Respuesta :

Answer:

part 1) [tex]13[/tex]

part 2) [tex]\frac{119}{45}[/tex]

part 3) [tex]2[/tex]

part 4) [tex]\frac{1,958,309}{128}[/tex]

part 5) [tex]4\ yd^2[/tex]

Step-by-step explanation:

The complete question in the attached figure

we know that

Applying PEMDAS

P ----> Parentheses first  

E -----> Exponents (Powers and Square Roots, etc.)  

MD ----> Multiplication and Division (left-to-right)  

AS ----> Addition and Subtraction (left-to-right)

Part 1) we have

[tex]\frac{2}{3}(6)+\frac{3}{4}(12)[/tex]

Remember that when multiply a fraction by a whole number, multiply the numerator of the fraction by the whole number and maintain the same denominator

so

[tex]\frac{12}{3}+\frac{36}{4}[/tex]

[tex]4+9=13[/tex]

Part 2) we have

[tex]2\frac{1}{3}(3\frac{2}{5}:3)[/tex]

Convert mixed number to an improper fraction

[tex]2\frac{1}{3}=2+\frac{1}{3}=\frac{2*3+1}{3}=\frac{7}{3}[/tex]

[tex]3\frac{2}{5}=3+\frac{2}{5}=\frac{3*5+2}{5}=\frac{17}{5}[/tex]

substitute

[tex]\frac{7}{3}(\frac{17}{5}:3)[/tex]

Solve the division in the parenthesis (applying PEMDAS)

[tex]\frac{7}{3}(\frac{17}{15})[/tex]

[tex]\frac{119}{45}[/tex]

Part 3) we have

[tex]\frac{7}{8}:(1\frac{1}{4}:4)[/tex]

Convert mixed number to an improper fraction

[tex]1\frac{1}{4}=1+\frac{1}{4}=\frac{1*4+1}{4}=\frac{5}{4}[/tex]

substitute

[tex]\frac{7}{8}:(\frac{5}{4}:4)[/tex]

Solve the division in the parenthesis (applying PEMDAS)

[tex]\frac{7}{8}:(\frac{5}{16})[/tex]

Multiply in cross

[tex]\frac{80}{40}=2[/tex]

Part 4) we have

[tex]18:(\frac{2}{3})^2+25:(\frac{2}{5})^7[/tex]

exponents first

[tex]18:(\frac{4}{9})+25:(\frac{128}{78,125})[/tex]

Solve the division

[tex](\frac{162}{4})+(\frac{1,953,125}{128})[/tex]

Find the LCD

LCD=128

so

[tex]\frac{32*162+1,953,125}{128}[/tex]

[tex]\frac{1,958,309}{128}[/tex]

Part 5) Find the area of triangle

The area of triangle is equal to

[tex]A=\frac{1}{2}(b)(h)[/tex]

substitute the given values

[tex]A=\frac{1}{2}(6)(1\frac{1}{3})[/tex]

Convert mixed number to an improper fraction

[tex]1\frac{1}{3}=1+\frac{1}{3}=\frac{1*3+1}{3}=\frac{4}{3}[/tex]

substitute

[tex]A=\frac{1}{2}(6)(\frac{4}{3})=4\ yd^2[/tex]

Ver imagen calculista