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3 years ago

3 feet of ribbons are needed for wrapping each present. How many presents can be wrapped with 102cm of ribbons?

2 answers:

8 0

1 imperial foot has approximately 30.48 metric centimeters, and for one present we need 3 ft, or namely 3(30.48) cm, how many can we get from 102 cm? 102 ÷ 3(30.48) ≈ 1.1155, so barely just one present.

kozerog [31]3 years ago

3 0

Answer:

1 present

Step-by-step explanation:

3 feet = 36 inches

1 inch = 2.54 cm, so 36 in = 36*2.54cm = 91.44cm

Each present is 91.44 cm, and 102 cm of ribbon can only be used to wrap one present.

The answer is 1 present

Hope this helps :)

Have a great day!

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3 years ago

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3 years ago

F(x)=-2(x-3)(x+5) convert quadratic functions to standard form

jasenka [17]

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3 years ago

A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of th

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so, when the boat was going up, it wasn't really going "b" fast, it was going " b - c " fast, because the current was eroding speed from it

now, when coming down, the return trip, well the length is the same, so the distance is also 100miles, it only took 2hrs though, because, the boat wasn't coming down "b" fast, it was coming down " b + c " fast, because the current was adding speed to it, so it came down quicker

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$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&100&b-c&5\\
downstream&100&b+c&2
\end{array}
\\\\\\
\begin{cases}
100=(b-c)5\\
\qquad \frac{100}{5}=b-c\\
\qquad 20=b-c\\
\qquad 20+c=\boxed{b}\\
100=(b+c)2\\
\qquad 50=b+c\\
\qquad 50=\boxed{20+c}+c
\end{cases}$

solve for "c", to see what's the current's speed

what's "b"? well 20+c = b

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3 years ago