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Ugo [173]
2 years ago
15

You’ve bought a cylindrical thermos 8 in. high with a base 6 in. across. How much coffee can you put in the thermos?

Mathematics
1 answer:
lana66690 [7]2 years ago
4 0

Answer:

3.706 liters

Step-by-step explanation:

Hello

assuming that these measures correspond to the internal measures of the thermos,

we can calculate the volume as, the volume of a cylinder, which is given by

V=\pi r^{2} h

where r is the radius of the base, and h the height

also

Diameter=2r

r=Diameter/2

Step 1

let

h=8 in

base = diameter = 6 inch

r=diameter/2

r=6/2 inch,r=3 inch

Step 2

Put the values into the equation

V=\pi r^{2} h\\V=\pi (3in)^{2} (8 in)\\V=\pi *72 \ in^{3} \\\\V=72\pi\ in^{3} \\\\

Step 3

Now, the unit (cubic inches) is not an usual unit for measuring coffee, let´s convert cubic inches into liters by knowing

1 inch = 2.54 cm

1 litro =  1000 cubic centimeters

if\ 1\ inch\ = 2.54\ cm\\(1\ inch)^{3} =(2.54\ cm)^{3}\\\\1\ inch^{3}=16.3870\ cm^{3}

1 cubic inch = 16.3870 cubic centimeters

1\ cubic\ inch\ = 16.3870\ cubic\ centimeters\\\\\\72\pi =x?\ cubic\ centimeters\\\\x=\frac{72\pi \cubic \ inch\ \*16.3870 }{1\ cubic\ inch} \\\\x=3706.65\ cubic\ centimeters

Step 4

Finally, convert cubic centimeters into liters1000\ cubic\ centimeters =1\ liter \\\\\\3706.65\ cubic\ centimeters = x?\ liters\ centimeters\\\\x=\frac{3706.65 \cubic \ centimeters\ \*1 liter}{1000 cubic centimeters} \\\\x=3706.65\ cubic\ centimeters\\x=3.706 Liters

I can put 3.706 liters in the thermos

Have a good day

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to compare the triangles, first we will determine the distances of each side

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</span>Solving 

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<span>AB = 6 units   BC = 11 units AC = 12.53 units
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<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

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</span><span>PQ = 5 units   QR = 9 units PR = 10.30 units</span> 
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<span>we have the <span>∆ABC   and the </span><span>∆MNO  </span><span> 
with all three sides equal</span> ---------> are congruent  
</span><span>we have the <span>∆JKL  </span>and the <span>∆PQR 
</span>with all three sides equal ---------> are congruent  </span>

 let's check

 Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

 1)     If ∆MNO   ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC 

<span> then </span>∆MNO<span> ≅</span> <span>∆ABC  </span> 

 a)      Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>M(-9, -4)----------------->  M1(-9,4)</span>

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

 b)      Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

<span>∆M1N1O1  M1(-9, 4), N1(-3, 4), and O1(-3, 15).</span>

<span>M1(-9, -4)----------------->  M2(9,4)</span>

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

 c)   Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

<span>∆M2N2O2  M2(9,4), N2(3,4), and O2(3, 15).</span>

<span>M2(9, 4)----------------->  M3(11,6)=A</span>

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

 ∆MNO  reflection------- >  ∆M1N1O1  reflection---- > ∆M2N2O2  translation -- --> ∆M3N3O3 

 The ∆M3N3O3=∆ABC 

<span>Therefore ∆MNO ≅ <span>∆ABC   - > </span>check list</span>

 2)     If ∆JKL  -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR 

<span> then </span>∆JKL ≅ <span>∆PQR  </span> 

 d)     Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

<span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).</span>

<span>J(17, -2)----------------->  J1(2,17)</span>

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

 e)      translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

<span>∆J1K1L1  J1(2, 17), K1(2, 12), and L1(-7, 12).</span>

<span>J1(2, 17)----------------->  J2(12,3)=P</span>

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

 ∆PQR  P(12, 3), Q(12, -2), and R(3, -2)

 ∆JKL  rotation------- >  ∆J1K1L1  translation -- --> ∆J2K2L2=∆PQR 

<span>Therefore ∆JKL ≅ <span>∆PQR   - > </span><span>check list</span></span>
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