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zimovet [89]
3 years ago
6

Natalie has $20 to spend on 5 sandwiches. Can she spend $4.39 on each sandwich? Explain.

Mathematics
2 answers:
snow_tiger [21]3 years ago
7 0
No.

First let's look at how much the sandwiches should cost at a maximum for a total of $20 for 5 sandwiches.

$20/5 = $4 per sandwich. So we can not spend over $4 per sandwich.

If you take the $4.39 and multiple it by the need of 5 sandwiches, you get $21.95. 

20 - 21.95 = -$1.95   -> Meaning you are over the budget by $1.95

OR

If you look at $20/$4.39 = 4.56 sandwiches 

And 4.56 sandwiches < 5 sandwiches.

uranmaximum [27]3 years ago
6 0
No she can't spend $4.39 for each sandwich because if you divide 20 by 5 your quotient will be 4.
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Hope this helps! I'm sorry for worse handwriting.

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Best explained and correct answer gets brainliest.
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U = (-2,3)

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How do I do this?<br>*Look at the directions in the photo*​
lora16 [44]

Answer:

Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2

Step-by-step explanation:

We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\

As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2

For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2

For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2

Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2

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Answer:

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