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3 years ago

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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.

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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.

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Degger [83]

Answer:

It is c: 15.625 cm³

Step-by-step explanation:

Formular for cube volumes:

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

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

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

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

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