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leonid [27]
3 years ago
12

Pierce currently has 10,000. What was the value of his money 5 years sgo

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
4 0
Where is he from bc that's very important
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“Number of teeth each child has in second grade class.”<br><br> Low or high variability?
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The answer is high variability
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Which graph represents (x - 1)2/9 (y-3)2/1
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Read 2 more answers
The isosceles triangle shown has a base of 4 y + 2 and a perimeter of 6 y + 12 .  Which expression represents the length of 1 o
liberstina [14]

Answer:

The expression that represents the length of 1 of the triangle's legs is y + 5

Step-by-step explanation:

An isosceles triangle has two sides equal which are the triangle legs. Let b represent the base of the triangle and l represent one of the triangle's legs. Then, the perimeter, P is given by

P = l + l + b

i.e P = 2l + b

From the question, P = 6y + 12 and b = 4y +2

∴ 6y + 12 = 2l + 4y + 2

6y - 4y + 12 - 2 = 2l

2y + 10 = 2l

∴ 2l = 2y + 10

Then,

l = (2y+10)/2

l = y + 5

Hence, the expression that represents the length of 1 of the triangle's legs is y + 5  

4 0
3 years ago
10 points help due tomarrow
strojnjashka [21]

Answer:

\boxed{0.2}

Step-by-step explanation:

Convert \dfrac{12}{60} to a decimal

Step 1. Reduce the fraction to its lowest terms.

Divide both numerator and denominator by their greatest common factor (12)

\dfrac{12}{60} =\dfrac{1}{5}

Step 2. Convert the denominator to a power of 10

Multiply both numerator and denominator by 2.

\dfrac{1}{5} \times \dfrac{2}{2} = \dfrac{2}{10}

Step 3. Divide the numerator by the denominator

Dividing by 10 moves the decimal point one place to the left.

\dfrac{2}{10} = \boxed{0.2}

4 0
3 years 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}

8 0
3 years ago
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