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Solnce55 [7]
3 years ago
15

The sum of a number and forty-six

Mathematics
1 answer:
ad-work [718]3 years ago
4 0

\textsf{Hey there!}

\mathsf{\star \ The\ sum\ of\ a\ number\ \&\  forty-six}

\frak{Let's\ lable\ the\ keypoints\ so\ it\ can\ be\ easier\ to\ solve}

\bullet \ \textsf{The word \bf{\underline{SUM}}}\textsf{\ means\ add/addition}}

\bullet\textsf{ \underline{"A number"} is an unknown number so we can lable it as \bf{x}}

\bullet\textsf{ \underline{forty-six} is 46}

\textsf{The sum (+) does in the middle while \underline{x} \& \underline{46} will either be on left/right}

- \ \textsf{We can say that x is on your left}

-\  \textsf{  + can be in your middle}

- \ \textsf{\& lastly 46 can be on your right}

\boxed{\textsf{Thus, your answer SHOULD LOOK like: \boxed{\huge\text{\bf{x + 46}}}}}\checkmark

\textsf{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

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he volume of a fish tank is 50 cubic feet. If the density is 0.2 fish over feet cubed, how many fish are in the tank?
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<em><u>Solution:</u></em>

Given that, volume of a tank is 50 cubic feet

Density is 0.2 fish over feet cubed

<em><u>To find: Number of fish in the tank</u></em>

Number of fish in the tank is found by multiplying the volume of tank and density

<em><u>The formula is given by:</u></em>

\text{Number of fish} = \tex{\text{volume of tank}} \times {\text{density of fish over feet cubed}}

<em><u>Substituting the given values, we get</u></em>

\text{Number of fish} = 50 ft^3 \times \frac{\text{ 0.2 fish}}{ft^3}\\\\ \text{Number of fish} =  50 \times 0.2 \text{ fish }\\\\ \text{Number of fish} = 10 \text{ fish }

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