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zzz [600]
3 years ago
13

Tom is buying topsoil for the flower bed shown below. One bag of topsoil covers 151515 square meters. How many bags of topsoil d

oes Tom need to cover his flower bed?
Mathematics
1 answer:
exis [7]3 years ago
6 0

Answer:

2 bags of topsoil.

Step-by-step explanation:

The attached figure shows a flower bed.

One bag of topsoil covers 15 square meters.

We need to find how many bags of topsoil does Tom need to cover his flower bed.

The area of the flower bed is :

A=\dfrac{1}{2}\times 12\times 5\\\\=30\ m^2  

Let he has to cover x bags of topsoil. So,

x = Area of flower bed/Area of 1 bag of top soil

x = 30/15

x = 2

Hence, he will need 2 bags of topsoil to cover his flower bed.

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Answer:

1/10 = 2/20

1/3 = 3/9

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Using Euler’s formula, how many edges does a polyhedron with 20 faces and 12 vertices have?
astra-53 [7]

Euler's formula for a polyhedron is given by:

\begin{gathered} F+V=E+2 \\ \text{where,} \\ F=\text{faces} \\ V=\text{vertices} \\ E=\text{edges} \end{gathered}

Make E, the subject of the formula:

\begin{gathered} F+V=E+2 \\ E=F+V-2 \end{gathered}

Put F = 20, V = 12, to obtain E,

\begin{gathered} E=20+12-2 \\ E=30 \end{gathered}

Therefore, there are 30 edges

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1 year ago
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}

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Answer:

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Step-by-step explanation:

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We move all terms containing x to the left, and all other terms to the right

6x=78

x=78/6

which should equal 13

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1 year ago
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