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Elenna [48]
3 years ago
11

denny can deliver 6 papers every 15 minutes and brendan can deliver 5 papers in that same amount of time. how many papers can th

ey deliver in an hour?
Mathematics
2 answers:
Evgesh-ka [11]3 years ago
7 0

So for this problem you know that 15 is 1/4 of an hour so to get the amount of papers each individual could deliver by themselves in an hour you would multiply the amount they can deliver in 15 minutes by 4. Denny delivers 6 so multiplying that by 4 is 24. Next Brendan delivers 5 in 15 minutes so multiplying that by 4 you get 20. The final step is to add 20 and 24 to get the total amount of papers between the two of them that gets delivered in an hour. This will result in your answer of 44.

marshall27 [118]3 years ago
3 0
30 in an hour if you times 6 x 5
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The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped
kakasveta [241]

Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

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(15 points)<br> NOTE: Angles not necessarily drawn to scale.
Akimi4 [234]
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Answer:

6.19=6\frac{19}{100}

Step-by-step explanation:

6.19=6\frac{19}{100}

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