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trapecia [35]
3 years ago
12

Tony wrote the following:

Mathematics
1 answer:
frutty [35]3 years ago
6 0

Answer:

yes, Tony’s statement correct

Step-by-step explanation:

Lets solve both Left hand side and right hand side separately and then equate.

LHS

7\frac{1}{4} -3\frac{3}{4} = \frac{29}{4} -\frac{15}{4}

= \frac{29}{4} -\frac{15}{4 } = \frac{14}{4}

Taking RHS

4\frac{1}{4} -\frac{3}{4} = \frac{17}{4} -\frac{3}{4} =\frac{14}{4}

clearly LHS = RHS hence both quantities are equal to each other

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Dennis_Churaev [7]

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Please need help on this
ser-zykov [4K]

Answer: In my answer I got 9 because the formula to find the area it would be A=1/2b*h



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Well we know that the base equals 6 we have heighth so we would have to do 1.5*6 and that would give us 9 but if we round it would be 10 because 9 is closer to 10 than it is to 0.


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3 years ago
A coffee seller makes a salary of $3,500 per month plus a $12 commission for every $100 of coffee he sells. A case of coffee cos
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Step-by-step explanation:

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3 years ago
(5x^3-8x^2+9x+12)/(x-3)<br><br> please answer with sentences on how to do it step by step. thanks!
Anastaziya [24]

Answer:

5x^2+7x+30+\frac{102}{x-3}

Step-by-step explanation:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}

6 0
3 years ago
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