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3 years ago

Determine between which consecutive integers the real zeros of each function are located f(x) = 3x3 – 10x2 + 22x – 4

1 answer:

7 0

we have to find which consective integer the real zero is present so for this we will assume
x=0:
f(0)= 3*0-10*0+22*0-4 = -4
now

 for: x=1:
f(1)=3*1-10*1+22*1-4=3-10+22-4=11
now it means that the real zero lies between 0 and 1

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Ivahew [28]

Answer:

$Width: \frac{3in}{8ft} = \frac{5in}{x} (cross multiply)$

= $40$ ÷3 ≈ 13.3ft

$Width: \frac {3in}{8ft} = \frac{6.5in}{x}$

= 17.3

$A=e$ × $w$

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2 years ago

You are planning to invest $5000 in an account earning 9% per year for retirement. a. If you put the $5000 in an account at age

ICE Princess25 [194]

Answer:

A).Amount = $218250

B). Amount = $88700

Step-by-step explanation:

A) .$5000 in an account at age 23, and withdraw it 42 years

Number of years t= 42 years

Principal P = $5000

Rate r= 9%

Number of times compounded n= 42

A= p(1+r/n)^(nt)

A= 5000(1+0.09/42)^(42*42)

A= 5000(1+0.002143)^(1764)

A= 5000(1.002143)^1764

A= 5000(43.65)

A= 218250

Amount = $218250

B).waits 10 years before making the deposit, so that it stays in the account for only 32 years

Number of years t= 32 years

Principal P = $5000

Rate r= 9%

Number of times compounded n= 32

A= p(1+r/n)^(nt)

A= 5000(1+0.09/32)^(32*32)

A= A= 5000(1+0.0028125)^(1024)

A= 5000(1.0028125)^1024

A= 5000(17.74)

A= 88700

Amount = $88700

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2 years ago

A fruit punch uses 1.5 cups of orange juice for every cup of apple juice.

timofeeve [1]

Are there any answer choices

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2 years ago

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A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

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3 years ago

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skelet666 [1.2K]

I think it would be -26

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2 years ago

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