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2 years ago

What two values of x are roots of the polynomial below

Mathematics

1 answer:

Anarel [89]2 years ago

6 0

Both A and B will be answers to the solution.

Since you already have the equation set and equal to 0, you can find the roots by first getting your a, b, and c values for the quadratic equation.

a = 1 (number attached to the x^2)
b = -3 (number attached to the x)
c = 5 (number not attached to a variable)

Then we can use these values in the quadratic equation.

$\frac{-b +/- \sqrt{ b^{2} -4ac } }{2a}$

$\frac{3 +/- \sqrt{ -3^{2} -4(1)(5) } }{2(1)}$

$\frac{3 +/- \sqrt{-11 } }{2}$

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This is an absolute value problem with LETTERS?! Absolute madness (no pun intended)

MAVERICK [17]

Answer:

Step-by-step explanation:

sjejeneissbebdejjesj

8 0

2 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

$2\times 1.5\times (r')^2=0.321$

$r'^2=0.107009$

$r'=0.327 m$

7 0

2 years ago

1. **What is the slope of a line that passes through the points (-1,30) and (4, 10)

tresset_1 [31]

Answer:

Slope = -4

Step-by-step explanation:

Given points are :

(-1,30) and (4, 10)

We need to find the slope of the line that passes through these points.

We know that, the formula for the slope of line is given by :

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have, x₁ = -1, y₁ = 30, x₂ = 4 and y₂ = 10

Put values in the above formula.

$m=\dfrac{10-30}{4-(-1)}\\\\=\dfrac{-20}{5}\\\\=-4$

So, the slope of the line is -4.

8 0

2 years ago

A cube has a side length of 14 meters. What is the volume of the cube?

Rudiy27

V=s^3
V=14^3
V=196*14
V=2744 cubic feet.

The answer is A.

4 0

2 years ago

Read 2 more answers

You are creating an open top box with a piece of cardboard that is 16 x 30“. What size of square should be cut out of each corne

Arada [10]

Answer:

$\frac{10}{3} \ inches$ of square should be cut out of each corner to create a box with the largest volume.

Step-by-step explanation:

Given: Dimension of cardboard= 16 x 30“.

As per the dimension given, we know Lenght is 30 inches and width is 16 inches. Also the cardboard has 4 corners which should be cut out.

Lets assume the cut out size of each corner be "x".

∴ Size of cardboard after 4 corner will be cut out is:

Length (l)= $30-2x$

Width (w)= $16-2x$

Height (h)= $x$

Now, finding the volume of box after 4 corner been cut out.

Formula; Volume (v)= $l\times w\times h$

Volume(v)= $(30-2x)\times (16-2x)\times x$

Using distributive property of multiplication

⇒ Volume(v)= $4x^{3} -92x^{2} +480x$

Next using differentiative method to find box largest volume, we will have $\frac{dv}{dx}= 0$

$\frac{d (4x^{3} -92x^{2} +480x)}{dx} = \frac{dv}{dx}$

Differentiating the value

⇒ $\frac{dv}{dx} = 12x^{2} -184x+480$

taking out 12 as common in the equation and subtituting the value.

⇒ $0= 12(x^{2} -\frac{46x}{3} +40)$

solving quadratic equation inside the parenthesis.

⇒ $12(x^{2} -12x-\frac{10x}{x} +40)$ =0

Dividing 12 on both side

⇒ $[x(x-12)-\frac{10}{3} (x-12)]$ = 0

We can again take common as (x-12).

⇒ $x(x-12)[x-\frac{10}{3} ]$ =0

∴ $(x-\frac{10}{3} ) (x-12)= 0$

We have two value for x, which is $12 and \frac{10}{3}$

12 is invalid as, w= $(16-2x)= 16-2\times 12$

∴ 24 inches can not be cut out of 16 inches width.

Hence, the cut out size from cardboard is $\frac{10}{3}\ inches$

Now, subtituting the value of x to find volume of the box.

Volume(v)= $(30-2x)\times (16-2x)\times x$

⇒ Volume(v)= $(30-2\times \frac{10}{3} )\times (16-2\times \frac{10}{3})\times \frac{10}{3}$

⇒ Volume(v)= $(30-\frac{20}{3} ) (16-\frac{20}{3}) (\frac{10}{3} )$

∴ Volume(v)= 725.93 inches³

6 0

2 years ago