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wariber [46]
2 years ago
12

A piece of metal has a mass of 2210 g. The volume of the metal is 260 cm³.

Mathematics
2 answers:
Mademuasel [1]2 years ago
7 0

Answer:

8.5g*cm^{-3}

Step-by-step explanation:

Density = mass / volume

\rho=\frac{m}{V}

\rho =\frac{2210}{260} =8.5g*cm^{-3}

Vlada [557]2 years ago
6 0

Answer:

density= 8.5\,\,\frac{g}{cm^3}

Step-by-step explanation:

Recall that the formula for density is:

density=\frac{mass}{volume}

therefore for this case we have:

density=\frac{mass}{volume} \\density=\frac{2210}{260} \,\frac{g}{cm^3} \\density= 8.5\,\,\frac{g}{cm^3}

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<u>Answer</u>:

The perimeter of rhombus WXYZ is 4 \sqrt{13}

<u>Step-by-step explanation:</u>

Step 1 :Finding length  of  XY

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Step 2 :Finding length  of  YZ

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here

x_1= 3

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Step 3 : :Finding length  of  ZW

Distance formula  = \sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}

here

x_1= 5

x_2=7

y_1= 5

y_2=2

ZW = \sqrt{(7-5)^2 +(5-2)^2}

ZW  = \sqrt{(2)^2 +(3)^2}

ZW  = \sqrt{4 +9}

ZW = \sqrt{(13)}

Step 4 :Finding length  of  WX

Distance formula  = \sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}

here

x_1= 7

x_2=5

y_1= 2

y_2= -1

WX = \sqrt{(7-5)^2 +((-1)-2)^2}

WX  = \sqrt{(2)^2 +(-3)^2}

WX  = \sqrt{4 +9}

WX = \sqrt{(13)}

Step 5: finding the perimeter of the rhombus

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