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Arte-miy333 [17]

3 years ago

6

You have a map of an area in France. The scale used is 2cm:8km. You want to ride to a national park. The park is shown on the ma

p as 16 cm away.
How far is that in kilometres?

1 answer:

slega [8]3 years ago

8 0

Divide the distance on the map by 2 to get the number of 8km segments there are, then multiply that by 8km for total distance.

16 cm / 2cm = 8

8 x 8km = 64km total.

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Please help is needed urgently! 8th grade math question?!

AURORKA [14]

ratio of the diagonal to the side of the rectangle is 17/8 or 17:8

Step-by-step explanation:

Length of Diagonal = 34 inches

Length of side = 16 inches

Base= 30 inches

We need to find the ratio of the diagonal to the side of the rectangle.

ratio of the diagonal to the side of the rectangle = $\frac{length of diagonal}{length of side}$

Putting values:

$Ratio=\frac{34}{16} \\Simplifying:\\Ratio=\frac{17}{8}$

So, ratio of the diagonal to the side of the rectangle is 17/8 or 17:8

Keywords: ratio

Learn more about ratio at:

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3 years ago

Please help me <br> Show your work <br> 10 points

Svet_ta [14]

<h2>Answer</h2>

After the dilation $\frac{5}{3}$ around the center of dilation (2, -2), our triangle will have coordinates:

$R'=(2,3)$

$S'=(2,-2)$

$T'=(-3,-2)$

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

$(x,y)$ → $(x-2, y+2)$

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor $\frac{5}{3}$ . Therefore our second partial rule will be:

$(x,y)$ → $\frac{5}{3} (x-2,y+2)$

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )$

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)$

$(x,y)$ → $(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

$R=(2,1)$

$R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

$R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})$

$R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})$

$R'=(2,3)$

$S=(2,-2)$

$S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

$S'=(2,-2)$

$T=(-1,-2)$

$T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

$T'=(-3,-2)$

Now we can finally draw our triangle:

8 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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