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xeze [42]
3 years ago
5

A cubical house can hold 4096 cm3 of air find the height of the house​

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

Answer:

The height is 16

Step-by-step explanation:

Given

Shape:  Cube

Volume = 4096cm^3

Required:

Determine the height of the house.

Volume is represented as:

Volume = L^3

Make L the subject

L = \sqrt[3]{Volume}

This gives:

L = \sqrt[3]{4096}

L= 16

The height is 16

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Y=f(x) = 8^x find f(x) when x= 1/3
lilavasa [31]

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2!!


4 0
3 years ago
Consider the triangle.<br> 45°<br> 80°<br> Find the measure of the unknown third angle.
Elina [12.6K]

Answer:

55 \degree

Step-by-step explanation:

Let the unknown angle be x

As the interior angles in a triangle are added up to 180°

x + 45 + 80 = 180 \\ x + 125 = 180 \\ x = 180 - 125 \\ x = 55 \degree

Hope this helps you.

Let me know if you have any other questions:-)

6 0
3 years ago
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How do I do this question? Please Help!
AURORKA [14]

Answer:

2x+50 and 5x-55 both are congruent or have same measure.

Step-by-step explanation:

Since we want to prove that both lines are parallel, this means no theorems that involve with parallel lines apply here.

First of, we know that AC is a straight line and has a measure as 180° via straight angle.

x+25 and 2x+50 are supplementary which means they both add up to 180°.

Sum of two measures form a straight line which has 180°.

Therefore:-

x+25+2x+50=180

Combine like terms:-

3x+75=180

Subtract 75 both sides:-

3x+75-75=180-75

3x=105

Divide both sides by 3.

x=35°

Thus, x = 35°

Then we substitute x = 35 in every angles/measures.

x+25 = 35°+25° = 60°

2x+50 = 2(35°)+50° = 70°+50° = 120°

5x-55 = 5(35°)-55 = 175°-55° = 120°

Since 2x+50 and 5x-55 have same measure or are congruent, this proves that both lines are parallel.

5 0
3 years ago
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What is the completely factored form of f(x)=x^3-7x^2+2x+4
xxMikexx [17]

Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)

Steps:

x^3-7x^2+2x+4

Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1},\\\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)\frac{x^3-7x^2+2x+4}{x-1}

\frac{x^3-7x^2+2x+4}{x-1}

\mathrm{Divide}\:\frac{x^3-7x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{x^3}{x},\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\\frac{x^3-7x^2+2x+4}{x-1}=x^2+\frac{-6x^2+2x+4}{x-1}

\mathrm{Divide}\:\frac{-6x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-6x^2}{x}=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\\frac{-6x^2+2x+4}{x-1}=-6x+\frac{-4x+4}{x-1}

\mathrm{Divide}\:\frac{-4x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-4x}{x}=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\\frac{-4x+4}{x-1}=-4

\mathrm{The\:Correct\:Answer\:is\:\left(x-1\right)\left(x^2-6x-4\right)}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

8 0
3 years ago
2/3+4/7 is it 26/21 or 6/21 or 3/5 or 6/10
Step2247 [10]

Answer:

26/21 = 1 5/20 = 4

5 0
3 years ago
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