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2 years ago

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According to the rational root theorem, the numbers below are some of the potential roots of f(x) = 10x3 29x2 – 66x 27. select a

ll that are actual roots.

1 answer:

exis [7]2 years ago

6 0

The actual roots of the function $f(x)=10x^{3}+29x^{2} -66x+27$ are -9/2, 3/5 and 1.

Given function $f(x)=10x^{3}+29x^{2} -66x+27$ .

Function is a relationship between two or more variables expressed in equal to form.

The roots of a polynomial function are the zeroes of the polynomial function. A polynomial function is a function that involves only non negative integer powers in an equation.

The polynomial function is given as:

$f(x)=10x^{3}+29x^{2} -66x+27$

factorize the above function

$f(x)=(2x+9)(5x-3)(x-1)$

Now put the function f(x) equal to zero.

f(x)=(2x+9)(5x-3)(x-1)

split the function means put all the expressions equal to zero as under:

(2x+9)(5x-3)(x-1)=0

solve each for the value of x

x=-9/2,x=3/5,x=1

Hence the roots of the function $f(x)=10x^{3} +29x^{2} -66x+27$ are which are also the values of x are -9/2,3/5,1.

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What number do we divide by -8 to get 6?

Marina86 [1]

-48

6* -8 = -48

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3 years ago

Read 2 more answers

Each ice cream cone at Izzy's Ice Cream Parlor contains $\frac{1}{32}$ of a gallon of ice cream. If Izzy has $\frac{3}{4}$ of a

harkovskaia [24]

Answer: 24 cones.

Step-by-step explanation:

We know that each cone contains 1/32 of a gallon of ice cream.

And Izzy has a total of 3/4 gallons.

How many cones we can make with this? This will be equal to the quotient between 3/4 gallons and 1/32 gallons, which will be equal to the number of times that we can fit 1/32 in 3/4

this is:

N = (3/4 gal)/(1/32 gal) = 24

This means that we can make 24 ice cream cones with 3/4 gal of ice cream.

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2 years ago

The radioactive isotope radium has a half-life of 1,600 years. The mass of a 100-gram sample of radium will be______ grams after

Olenka [21]

To solve this we are going to use the half life equation $N(t)=N_{0} e^{( \frac{-0.693t}{t _{1/2} }) }$
Where:
$N_{0}$ is the initial sample
$t$ is the time in years
$t_{1/2}$ is the half life of the substance
$N(t)$ is the remainder quantity after $t$ years

From the problem we know that:
$N_{0} =100$
$t=200$
$t_{1/2} =1600$

Lets replace those values in our equation to find $N(t)$ :
$N(200) =100e^{( \frac{(-0.693)(200)}{1600}) }$
$N(200)=100e^{( \frac{-138.6}{1600} )}$
$N(200)=100e^{-0.086625}$
$N(200)=91.7$

We can conclude that after 1600 years of radioactive decay, the mass of the 100-gram sample will be 91.7 grams.

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3 years ago

What is the correct classification of the system of equations below? 14x + 2y = 10 y + 7x = -5 parallel coincident intersecting

marissa [1.9K]

In standard form the two equations are
7x +y = 5
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The equations describe lines that are parallel.

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3 years ago

The teacher asked 4 students to measure 4 different objects. The table shown represents their measurements. Which student had th

Maru [420]

Student 4 has the largest percent error.

Step-by-step explanation:

The percent error of a measurement is given by

$p=\frac{|x-x_{true}|}{x_{true}}\cdot 100$

where

x is the value of the measurement

$x_{true}$ is the actual value

For student 1:

$x=49\\x_{true}=50$

Percent error:

$p=\frac{|49-50|}{50}\cdot 100=0.02\cdot 100=2\%$

For student 2:

$x=25.5\\x_{true}=25$

Percent error:

$p=\frac{|25.5-25|}{25.5}\cdot 100=0.0196\cdot 100 = 1.96\%$

For student 3:

$x=7.5\\x_{true}=8$

Percent error:

$p=\frac{|7.5-8|}{8}\cdot 100=0.0625\cdot 100 = 6.25\%$

For student 4:

$x=2.1\\x_{true}=1.6$

Percent error:

$p=\frac{|2.1-1.6|}{1.6}\cdot 100=0.3125\cdot 100 = 31.25\%$

So, student 4 has the largest percent error.

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4 0

3 years ago