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viktelen [127]
3 years ago
7

7

Mathematics
1 answer:
lbvjy [14]3 years ago
3 0

The density of liquid C at the given mass and volume is 1.6 g/cm³.

The given parameters:

  • Density of liquid A, = 1.8 g/cm³
  • Density of liquid B, = 1.2 g/cm³
  • Volume of liquid C, = 120 cm³

<h3>Mass of liquid C</h3>

The mass of liquid C is calculated as follows;

mass =density \times volume\\\\mass _{C} = (1.8 \times 80 ) + (1.2 \times 40)\\\\mass _{C} = 192 \ g

<h3>The density of liquid C</h3>

The density of liquid C is calculated as follows;

Density = \frac{mass}{volume} \\\\Density_{C} =\frac{192 \ g}{120 \ cm^3} = 1.6 \ g/cm^3

Learn more about density here: brainly.com/question/6838128

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Write a two-column proof. Given: line BD bisects angle CBE. Prove: angle ABD approximately equal to angle FBD.
ale4655 [162]

Answer:

The proof is derived from the summarily following equations;

∠FBE + ∠EBD = ∠CBA + ∠CBD

∠FBE + ∠EBD = ∠FBD

∠CBA + ∠CBD = ∠ABD

Therefore;

∠ABD ≅ ∠FBD

Step-by-step explanation:

The two column proof is given as follows;

Statement {}                                       Reason          

\underset{BD}{\rightarrow} bisects ∠CBE {}                            Given

Therefore;

∠EBD ≅ ∠CBD  {}                              Definition of angle bisector

∠FBE ≅ ∠CBA {}                                Vertically opposite angles are congruent

Therefore, we have;

∠FBE + ∠EBD = ∠CBA + ∠CBD {}     Transitive property

∠FBE + ∠EBD = ∠FBD {}                    Angle addition postulate

∠CBA + ∠CBD = ∠ABD {}                   Angle addition postulate

Therefore;

∠ABD ≅ ∠FBD        {}                          Transitive property.

3 0
3 years ago
Find the volume of this cube in cm³​
Degger [83]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>c</u><u>:</u><u> </u><u>1</u><u>5</u><u>.</u><u>6</u><u>2</u><u>5</u><u> </u><u>cm³</u>

Step-by-step explanation:

Formular for cube volumes:

{ \tt{volume = (side \times side \times side)}} \\ { \tt{v = (s_{1} \times s_{2} \times s_{3}) }}

since sides of a cube are identical:

{ \sf{s_{1} = {s_{2} =  {s_{3}}}}}

Therefore:

{ \sf{volume = (250 \times 250 \times 250) \: mm {}^{3} }} \\  { \sf{v =15,625,000 \:  {mm}^{3}  }}

but 1 mm³ = 0.000001 cm³

{ \sf{v = (15,625,000 \times  \{1 \times  {10}^{ - 6} \} ) \: cm}} \\ { \sf{v = 15.625 \:  {cm}^{3} }}

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