Question

Calculate 51 + ( 6 + 3 x 2 ) ÷ 3

Answer 1

Answer:

a) 55 b)=21/32 c)x=6

Step-by-step explanation:

a)Follow the PEMDAS order of operations

Calculate within parentheses

(6 +3 x 2)

(6+6)

= 12

=51+ 12 ÷ 3

Multiply and divide (left to right)

12÷ 3=4

51 +4

Add and subtract (left to right)

51 + 4=55

B)

Follow the PEMDAS order of operations

29 – 8) ÷ 32

Calculate within parentheses

(29 – 8) =21

= 21 ÷32

Multiply and divide (left to right)

21 ÷32

=21/1× 1/32

=21/32

C)let the unknown be X

6+ 5× X- 16 = 20

Group like terms

5x +6- 16 = 20

Add/Subtract the numbers:

6- 16 = -10

5x - 10= 20

Add 10 to both sides

5x- 10 + 10 =20 + 10

Simplify

5x=30

Divide both sides by 5

5x/5=30/5

x=6

Answer 2

$\huge\text{Hey there!}$

$\mathsf{51 + ( 6 + 3 \times 2 ) \div 3}\\\\\mathsf{= 51 + \dfrac{6+ 3\times2}{3}}\\\\\mathsf{= 51 + \dfrac{6 + 6}{3}}\\\\\mathsf{= 51 + \dfrac{12}{3}}\\\\\mathsf{= 51 + 4}\\\\\mathsf{= 55}\\\\\huge\text{Therefore, the answer is: \boxed{\mathsf{55}}}\huge\checkmark$

$\mathsf{(29 - 8) \div 32}\\\\\mathsf{= \dfrac{29 - 8}{32}}\\\\\mathsf{= \dfrac{29 - 8}{32 - 0}}\\\\\mathsf{= \dfrac{21}{32}}\\\\\mathsf{= 21\div 32}\\\\\mathsf{= 0.65625}\\\\\huge\text{Therefore, the answer is: \boxed{\mathsf{0.65625}}}\huge\checkmark$

$\mathsf{6 + 5 \times p - 16 = 20}\\\mathsf{\rightarrow 6 + 5p - 16 = 20}\\\mathsf{\rightarrow 5p - 10 = 20}\\\text{ADD 10 to BOTH SIDES}\\\mathsf{5p - 10 + 10 = 20 + 10}\\\text{SIMPLIFY IT!}\\\mathsf{5p = 20 + 10}\\\mathsf{5p = 30}\\\text{DIVIDE 5 to BOTH SIDES}\\\mathsf{\dfrac{5p}{5}= \dfrac{30}{5}}\\\text{SIMPLIFY IT!}\\\mathsf{p = \dfrac{30}{5}}\\\mathsf{p = 6}\\\\\huge\text{Therefore, the answer is: \boxed{{\mathsf{6}}}}\huge\checkmark$

$\large\textsf{Lets check to make sure \#3 is correct!}\\\mathsf{6 + 5\times \bold{6} - 16}\\\mathsf{= 6 + 30 - 16}\\\mathsf{= 36 - 16}\\\mathsf{= 20}\\\\\huge\textsf{Thus, the answer in the blank IS: \boxed{\mathsf{6}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~$\frak{Amphitrite1040:)}$