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OleMash [197]
2 years ago
8

Hi! Could you help me me out and solve these two problems? "eight less than the product of seven and xxx", and "the sum of six a

nd the product of three and d". Thank you!
Mathematics
1 answer:
Ann [662]2 years ago
3 0

Given:

The given statement are:

a. Eight less than the product of seven and x.

b. The sum of six and the product of three and d.

To find:

The expression for the given statements.

Solution:

a.

Product of 7 and x is 7×x = 7x.

Eight less than the product of seven and x is 7x - 8.

Therefore, the required expression for the statement "Eight less than the product of seven and x" is 7x-8.

b.

Product of 3 and d is 3×d = 3d.

The sum of six and the product of three and d is 6+3d.

Therefore, the required expression for the statement "The sum of six and the product of three and d" is 6+3d.

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4 0
3 years ago
Linda can bicycle 48 miles in the same time as it takes her to walk 12 miles. She can ride 9 mph faster than she can walk. How f
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Answer:

\frac{48}{r+9}=\frac{12}{r}

Step-by-step explanation:

Let r represent Linda's walking rate.                      

We have been given that Linda can ride 9 mph faster than she can walk, so Linda's bike riding rate would be t+9 miles per hour.

\text{Time}=\frac{\text{Distance}}{\text{Rate}}

We have been given that Linda can bicycle 48 miles in the same time as it takes her to walk 12 miles.

\text{Time while riding}=\frac{48}{r+9}

\text{Time taken while walking}=\frac{12}{r}

Since both times are equal, so we will get:

\frac{48}{r+9}=\frac{12}{r}

Therefore, the equation \frac{48}{r+9}=\frac{12}{r} can be used to solve the rates for given problem.

Cross multiply:

48r=12r+108

48r-12r=12r-12r+108

36r=108

\frac{36r}{36}=\frac{108}{36}

r=3

Therefore, Linda's walking at a rate of 3 miles per hour.

Linda's bike riding rate would be t+9\Rightarrow 3+9=12 miles per hour.

Therefore, Linda's riding the bike at a rate of 12 miles per hour.

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