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3 years ago

How do I solve this operation

Mathematics

1 answer:

ivann1987 [24]3 years ago

6 0

4 / ( 3 + 5i ) = [ 4( 3 - 5i)] / [ 3^2 - (5i)^2 ] = [ 4( 3 - 5i)] / 34 = [ 2( 3 - 5i)] / 17 =
6 / 17 - 10i / 17;

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An investor invested a total of $2500 in two mutual funds. One fund earned a 5% profit while the other earned a 3% profit. If th

7nadin3 [17]

Answer:

$500 and $2000

Step-by-step explanation:

Let x represent the total investment = $2500

also, this total is split into two different funds

Lets represent these funds as a and b, such that fund a yields a profit of 3% and fund b yield a profit of 5%

So,

a + b = x

a + b = 2500 ......eq 1

Profit from each fund gives;

0.03 a + 0.05b = 115 ....eq 2

Solve simultaneously using substitution method

From eq 1;

b = 2500 - a

Slot in this value in eq 2

0.03a + 0.05 (2500 - a) = 115

expand

0.03a + 125 - 0.05a = 115

collect like terms

0.03a - 0.05a = 115 - 125

-0.02a = -10

Divide both sides by -0.02

a = $500

Put this value of a in eq 1

500 + b = 2500

Subtract 500 from both sides

b = 2500 - 500

b = $2000

3 0

3 years ago

Please help quickly Evaluate the expression: -|5+(2-8)|+|7(2-8)|

DanielleElmas [232]

$\displaystyle\\-\Big|5+(2-8)\Big|+\Big|7(2-8)\Big|=\\\\=-\Big|5+(-6)\Big|+\Big|7\times(-6)\Big|=\\\\=-\Big|-1\Big|+\Big|-42\Big|=\\\\=-(+1)+42=\\\\=-1+42=\boxed{\bf41}$

4 0

2 years ago

Read 2 more answers

What is the completely factored form of f(x)=x^3-7x^2+2x+4

xxMikexx [17]

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

$Steps:$

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

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

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

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

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

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