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Last weekend Ali visited an aviation festival near his neighborhood. In an experiment at the festival, an engineer introduced the element rhenium, which is used to make jet engine parts. The engineer showed the onlookers a small pyramid-shaped part of the element. The part was obtained by cutting one corner of a cube with edges of length $6$ centimeters as shown.

Ali read on a scale that the part weighs $94.5$ grams.

What is the density of rhenium?Ali wonders.

When two substances are mixed, the density of the mixture will still be the ratio of total mass to total volume. A carpenter uses four identical prisms with right triangle bases to create a cube-shaped mold as shown in the applet.
### Hint

### Solution

a Calculate the volume of the hollow part.

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b The carpenter mixes $52.5$ grams of glue and $27.5$ grams of sawdust, and pours it into the mold. What is the density of the mixture? If necessary, round the answer to two decimal places.

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a Use trigonometric ratios to find the dimensions of the mold.

b Use the volume of the hollow and the mass of the mixture.

a Consider the top view of the mold.
Using the cosine ratio of $37_{∘},$ the hypotenuse $h$ can be found.
Now, the sine ratio of $37_{∘}$ can be used to calculate the length $b$ of the opposite side.
Since identical triangular prisms are used, the other prisms' bases have the same side lengths.

$sin37_{∘}=HypotenuseOpposite side $

SubstituteValues

Substitute values

$sin37_{∘}=5b $

Solve for $b$

MultEqn

$LHS⋅5=RHS⋅5$

$5⋅sin37_{∘}=b$

RearrangeEqn

Rearrange equation

$b=5⋅sin37_{∘}$

UseCalc

Use a calculator

$b=3.009075…$

RoundInt

Round to nearest integer

$b≈3$

As the diagram indicates, the base of the hollow part is a square with a side length of $5$ centimeters. Additionally, the mold is a cube with an edge length of $7$ centimeters. Therefore, the hollow is a square prism with a base edge length of $5$ centimeters and height of $7$ centimeters.

The volume of this square prism can be calculated as the product of its base area and its height. Since the base of the prism is a square with a side length of $5$ centimeters, the base area is $5_{2}$ square centimeters. The volume of the hollow is $175cm_{3}.$ b Start by finding the mass of the mixture that the carpenter prepared.

$Glue52.5 + Sawdust27.5 = Total80 $

This amount of mixture occupies a volume of $175$ cubed centimeters. Since density is a measure of mass per volume, the density of the mixture is equal to the quotient of the mass and volume. The density of the mixture is about $0.46g/cm_{3}.$ After discussing hexagonal cells of beehives, Diego thinks he can approximate the number of cells in his body. His teacher recommends for Diego to treat a cell like a sphere with a diameter of $2×10_{-3}$ centimeters.

If Diego's weight is $60$ kilograms and the density of a cell is approximately the density of water, which is $1$ gram per cubic centimeter, help Diego approximate the number of cells in his body. Write the answer in scientific notation.

About $1.4×10_{13}$ cells

The formula for the volume of a sphere is $V=34 πr_{3},$ where $r$ is the radius of the sphere.

To find the number of cells $N,$ Diego's weight $M$ should be divided by the mass of a cell $m.$
### Finding the Volume of a Cell

With knowledge of the diameter of a cell, its radius can be calculated by dividing the diameter by $2.$
The volume of a cell is about $4.2×10_{-9}$ cubic centimeters. ### Finding the Mass of a Cell

The density of a cell is $1g/cm_{3}$ and its volume is $4.2×10_{-9}cm_{3}.$ By multiplying these values, the mass of a cell can be found.
### Finding the Number of Cells

The mass of a cell was found in grams. Therefore, Diego's weight should also be written in grams. To do so, use the conversion factor of $1000kgg .$
The number of cells is approximately $1.4×10_{13}.$

$N=mM $

Recall that the density of a substance is equal to its mass divided by its volume. In other words, the mass of a substance is its density times its volume. $d=Vm ⇔m=d⋅V $

Since the density of a cell is given, the volume of a cell can be calculated first. $22×10_{-3} =10_{-3}cm $

Now, use the formula for the volume of a sphere.
$V=34 πr_{3}$

Substitute

$r=10_{-3}$

$V=34 π(10_{-3})_{3}$

Evaluate right-hand side

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$V=34 π(10_{-9})$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$V=34π (10_{-9})$

UseCalc

Use a calculator

$V=(4.188790…)(10_{-9})$

RoundDec

Round to $1$ decimal place(s)

$V≈(4.2)(10_{-9})$

Rewrite

Rewrite $(4.2)(10_{-9})$ as $4.2×10_{-9}$

$V≈4.2×10_{-9}$

$1cm_{3} g ⋅4.2×10_{-9}cm_{3} =4.2×10_{-9}g $

$60kg ⋅1000kg g =60000g $

Finally, substitute the values into the formula mentioned at the beginning to calculate the number of cells.
$N=mM $

SubstituteII

$M=60000g$, $m=4.2×10_{-9}g$

$N=4.2×10_{-9}g60000g $

Evaluate right-hand side

CrossCommonFac

Cross out common factors

$N=4.2×10_{-9}g 60000g $

SimpQuot

Simplify quotient

$N=4.2×10_{-9}60000 $

WriteProdFrac

Write as a product of fractions

$N=4.260000 (10_{-9}1 )$

FracToNegExponent

$a_{m}1 =a_{-m}$

$N=4.260000 (10_{9})$

CalcQuot

Calculate quotient

$N=(14285.714285714…)(10_{9})$

Multiply

Multiply

$N=14285714285714$

RoundSigDig

\RoundSigDig{2}

$N≈14000000000000$

Write in scientific notation

$N≈1.4×10_{13}$

Throughout the lesson, different cases involving concepts of density based on area and volume have been discussed. Considering these situations, the challenge presented at the beginning of the lesson can be solved.

Ali knows that the small pyramid-shaped part of the element was obtained by cutting one corner of a cube with edges of length $6$ centimeters.

Given that the small piece's weight is $94.5$ grams, the density of rhenium can be calculated.

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The volume of a pyramid is one-third of the product of its base area and height.

From the diagram, it can be seen that the pyramid has three congruent edges, each is $3$ centimeters in length.

Additionally, the pyramid's base is an isosceles right triangle. Therefore, its area is half the product of its legs.

$B=21 bh⇓B=21 (3)(3)=4.5 $

Recall that the volume of a pyramid is one-third of the product of its base area and height. Using the fact that the height of the pyramid is $3$ centimeters, its volume can be determined.
Therefore, $94.5$ grams of rhenium has a volume of $4.5$ cubic centimeters. Now, using the definition of density, the volume of rhenium can be calculated.
The density of rhenium is $21$ grams per cubic centimeter.