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garik1379 [7]
3 years ago
15

Which interval is the solution set to 0.35x – 4.8 < 5.2 – 0.9x?

Mathematics
2 answers:
Allushta [10]3 years ago
7 0
No intervals are showing, but here is how to solve this equation.

0.35x - 4.8 < 5.2 - 0.9x
Add 4.8 to both sides
0.35x < 10 - 0.9x
add 0.9x to both sides
1.25x < 10
divide both sides by 1.25
x < 8

Interval Notation:
If we graphed this on a number line, x goes toward negative infinity.

Answer: (-∞, 8)

Hope this helps! :)
Nuetrik [128]3 years ago
4 0

Answer:Answer: (-∞, 8)

Step-by-step explanation:

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HELP i’m having trouble with my homework assignments
Aleksandr-060686 [28]

Answer:

Collin: about $401 thousand

Cameron: about $689 thousand

Step-by-step explanation:

A situation in which doubling time is constant is a situation that can be modeled by an exponential function. Here, you're given an exponential function, though you're not told what the variables mean. That function is ...

P(t)=P_0(2^{t/d})

In this context, P0 is the initial salary, t is years, and d is the doubling time in years. The function gives P(t), the salary after t years. In this problem, the value of t we're concerned with is the difference between age 22 and age 65, that is, 43 years.

In Collin's case, we have ...

P0 = 55,000, t = 43, d = 15

so his salary at retirement is ...

P(43) = $55,000(2^(43/15)) ≈ $401,157.89

In Cameron's case, we have ...

P0 = 35,000, t = 43, d = 10

so his salary at retirement is ...

P(43) = $35,000(2^(43/10)) ≈ $689,440.87

___

Sometimes we like to see these equations in a form with "e" as the base of the exponential. That form is ...

P(t)=P_{0}e^{kt}

If we compare this equation to the one above, we find the growth factors to be ...

2^(t/d) = e^(kt)

Factoring out the exponent of t, we find ...

(2^(1/d))^t = (e^k)^t

That is, ...

2^(1/d) = e^k . . . . . match the bases of the exponential terms

(1/d)ln(2) = k . . . . . take the natural log of both sides

So, in Collin's case, the equation for his salary growth is

k = ln(2)/15 ≈ 0.046210

P(t) = 55,000e^(0.046210t)

and in Cameron's case, ...

k = ln(2)/10 ≈ 0.069315

P(t) = 35,000e^(0.069315t)

5 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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