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defon
3 years ago
15

Help please :) about to post more too

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
5 0

Answer:

Lets say test tubes = t, and beakers = b

1 pack of (t) is $4 less than 1 pack of (b)

Since i have no prior information we are going to use variables for this equation:

1t (1 pack of test tubes) is $4 less than 1b (1 set of beakers)

so to quantify the equation, we have 8t and 12b.

if b is a number that IS quantifiable such as $5 we can easily figure out this answer.

Lets use and example that 1 set of beakers is $8, if we multiply $8 by 12 (the number of sets of beakers), we get: 96

Using the same example, if 1t is $4 less than 1b than 1t = $4. So, if we multiply $4 by 8 (the amount of packs of test tubes), we get: 32

If you take both of those numbers: 96, and 32 and you divide them you get 3. so that means that 1t = 3b

Answer = 1t = 3b

This may not be correct due to the little information that i got however i hope that, that works out for you :)

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7 0
3 years ago
A conical vessels with base radius 5 cm and height 24cm is full of water this water is emptied into a cylindrical vessel of base
Aliun [14]

Answer:

2\ cm

Step-by-step explanation:

We\ are\ given\ that,\\Radius\ of\ the\ conical\ vessel=5\ cm\\Height\ of\ the\ conical\ vessel=24\ cm\\Hence,\\As\ we\ know\ that,\\Volume\ of\ a\ cone=\frac{1}{3}\pi r^2h\\Hence,\\The\ volume\ of\ the\ conical\ vessel=\frac{1}{3}\pi (5)^224=  \frac{1}{3}\pi 25*24=200\pi \\Hence,\\The\ volume\ of\ the\ conical\ vessel=The\ amount\ of\ water\ filled\ in\ it\\Hence,\\The\ amount\ of\ water\ filled\ in\ the\ conical\ vessel=200\pi\\Lets\ consider\ the\ cylindrical\ vessel:\\Volume\ of\ a\ cylinder=\pi r^2h

As\ we\ are\ not\ said\ that\ the\ cylindrical\ vessel\ was\ empty,\ let\ its\ initial\\ height\ be\ h_1\\Radius\ of\ the\ Cylindrical\ vessel=10\ cm\\Initial\ height\ of\ the\ cylindrical\ vessel=h_1\\Initial\ volume\ of\ the\ cylindrical\ vessel=\pi *10^2*h_1=100h_1\pi \\Initial\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi \\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=Final\ amount\ of\ water\\ in\ the\ cylindrical\ vessel +Amount\ of\ water\ in\ conical\ vessel

Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi +200\pi \\Now,\\Let\ the\ final\ height\ be\ h_2\\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=\pi *10^2*h_2=100h_2\pi \\Hence,\\100h_2\pi =100h_1\pi +200\pi \\Hence,\\100h_2=100h_1+200\\100h_2-100h_1=200\\100(h_2-h_1)=200\\h_2-h_1=2\\Hence,\\As\ the\ rise\ in\ level\ is\ the\ difference\ between\ the\ initial\ and\ final\\ heights:\\The\ rise\ in\ level\ of\ water=2\ cm

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