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3 years ago

H(x) = x4+ 2x3- 10x2 -18x+9

2 answers:

5 0

Let's group the cubic function: $h(x)=x^4+2x^3-10x^2-18x+9=(x-3)(x+3)(x^2+2x-1)$ .

The first two roots are $\pm3$ however to find the last two roots we need to solve the square equation:

$x^2+2x-1=0
\\D=b^2-4ac=2^2-4\cdot1\cdot(-1)=8$ . Now we know the discriminator, using the discriminator we're able to find the roots of the equation: $x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}=\frac{-2\pm\sqrt{8}}{2}=\frac{-2\pm2\sqrt{2}}{2}=\frac{2(\sqrt{2}\pm1)}{2}$ . The roots of the square equation are $x_{1}=-1-\sqrt{2}$ and $x_{2}=\sqrt{2}-1$ .

The roots of the cubic function are $\pm3$ , $-1-\sqrt{2}$ and $\sqrt{2}-1$ .

Tamiku [17]3 years ago

3 0

H (x) = (x+3)(x-3)(2x+x^{2} -1)

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3 years ago

What is the area of a regular hexagon (6 sides) whose

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Step-by-step explanation:

With reference to the regular hexagon, from the image above we can see that it is formed by six triangles whose sides are two circle's radii and the hexagon's side. The angle of each of these triangles' vertex that is in the circle center is equal to 360∘6=60∘ and so must be the two other angles formed with the triangle's base to each one of the radii: so these triangles are equilateral.

The apothem divides equally each one of the equilateral triangles in two right triangles whose sides are circle's radius, apothem and half of the hexagon's side. Since the apothem forms a right angle with the hexagon's side and since the hexagon's side forms 60∘ with a circle's radius with an endpoint in common with the hexagon's side, we can determine the side in this fashion:

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3 years ago

The product of 3001 x 25 is ?

Nookie1986 [14]

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3 years ago

Read 2 more answers

The diameter and height of a right circular cylinder are found at a certain instant to be 10 centimeters and 20 centimeters, res

Andrej [43]

Answer:

Step-by-step explanation:

Here's the formula for the volume of a right circular cylinder:

$V=\pi r^2h$

Here's what we are given and what we need to find:

Given that d = 10 cm, h = 20 cm, dd/dt = 1 cm/sec

Need to find dh/dt when V is constant

Since our formula has a radius in it and not a diameter but the info given is a diameter, we can use the substitution that

$2r =d$ so

$r=\frac{1}{2}d$

Now we can rewrite the formula in terms of diameter:

$V=\pi (\frac{1}{2}d)^2h$ which simplifies down to

$V=\frac{\pi }{4}d^2h$

Now we will take the derivative of this equation with respect to time using the product rule. That derivative is

$\frac{dV}{dt}=\frac{\pi }{4}[d^2*\frac{dh}{dt}+2d\frac{dd}{dt}*h]$

Now we can fill in our values. Keep in mind that if the volume is constant, there is no change in the volume, so dV/dt = 0.

$0=\frac{\pi }{4}[(100\frac{dh}{dt}+2(10)(1)(20)]$ and

$0=\frac{\pi }{4}(100\frac{dh}{dt}+400)$

Multiply both sides by pi/4 to get

$0=100\frac{dh}{dt}+400$ and solve for dh/dt:

$-4=\frac{dh}{dt}$

Interpreted within the context of our problem, this means that the volume will be constant at those given values of diameter and height when the liquid in the cylinder is dropping at a rate of 4 cm/sec.

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3 years ago