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Mamont248 [21]
4 years ago
15

What is 7 and 25 over 18

Mathematics
1 answer:
grandymaker [24]4 years ago
8 0
7+25 / 18
32/18
or
1 14/18
or
1 7/9

Answer: 32/18 or 1 14/18 or 1 7/9
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From 9 P.M. to 5 A.M. the temperature dropped 20°, making the rate over 8 hours, -2.5° Per Hour. By 5 A.M., the temperature is -5° since we had to do 15 - 20. From 5 AM to 5 P.M. is 12 hours so -2.5 x 12 = -30 making 30 how much the temperature dropped between 5 A.M. to 5 P.M. Finally, our last equation is -5 - 30 making -35 our answer.
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How would u write Subtract 2 from 5 times d​
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5D-2

Step-by-step explanation:

5 times d = 5d. subtract 2 is after the 5d

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What is the answer to 5 x 2 x 2 + 2?
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3 years ago
Perform the indicated operation. Be sure the answer is reduced.
avanturin [10]
<h3>Given Equation:-</h3>

\boxed{ \rm  \frac{4x^{2}y^{3}z}{9} \times  \frac{45y}{8 {x}^{5} {z}^{5} }}

<h3>Step by step expansion:</h3>

\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times  \dfrac{45y}{8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times  \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times  \dfrac{45y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times  \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}z}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5y \times  {y}^{3} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5y {}^{0}  \times  {y}^{3 + 1} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5 \times  {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \bf  \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}

8 0
2 years ago
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