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3 years ago

Find all the zeros of the equation -2x^4+27x^2+1200=0

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

4 0

For this case we have the following polynomial:
$-2x ^ 4 + 27x ^ 2 + 1200 = 0$
To solve the problem we apply the following steps:
1) We make a change of variables u = x ^ 2
2) We rewrite the polynomial
3) Solve the quadratic equation using the formula for the quadratic equation
4) We return the change and find four roots
Answer:
See attached image for procedure and solution

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The product of (n2 – 6n + 3) and -4n is________ When this product is multiplied by -n, the result is________

krek1111 [17]

-4n^3+24n^2-12n, 4n^4-24n^3+12n^2

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3 years ago

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Newton’s law of cooling states that dx/dt= −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k

Vadim26 [7]

Answer:

(a) The solution to the differential equation is x = A_0Coswt + Ce^(-kx)

(b) The initial condition t > 0 will not make much of a difference.

Step-by-step explanation:

Given the differential equation

dx/dt= −k(x − A); t > 0, A = A_0Coswt

(a) To solve the differential equation, first separate the variables.

dx/(x - A) = -kdt

Integrating both sides, we have

ln(x - A) = -kt + c

x - A = Ce^(-kt) (Where C = ce^(-kt))

x = A + Ce^(-kx)

Now, we put A = A_0Coswt

x = A_0Coswt + Ce^(-kx) (Where C is constant.)

And we have the solution.

(b) Since temperature t ≠ 0, the initial condition t > 0 will not make much of a difference because, Cos(wt) = Cos(-wt).

It is not any different from when t < 0.

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3 years ago

angel's dad bought a television for her completing all of her digital learning day lesson. he want to warp the gift. the box mea

atroni [7]

Answer: 13 sheets of paper

Step-by-step explanation:

We are given the dimensions of the box and the wrap paper:

Box:

$(56 in)(8 in)(36 in)$

Warp paper:

$(26 in)(17 in)$

Now we need to find the surface area of the box and the area of the wrap paper:

Box:

$surface_{1}=(56 in)(8 in)(2)=896 in^{2}$

$surface_{2}=(8 in)(36 in)(2)=576 in^{2}$

$surface_{3}=(56 in)(36 in)(2)=4032 in^{2}$

$surface-area=896 in^{2}+576 in^{2}+4032 in^{2}=5504 in^{2}$

Warp paper:

$area=(26 in)(17 in)=442 in^{2}$

Dividing the area of the box by the area of the paper:

$\frac{surface-area}{area}=\frac{5504 in^{2}}{442 in^{2}}=12.45 \approx 13$

This means Angel's dad needs to purchase 13 sheets of wrapping paper.

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3 years ago

The dimensions of a triangular prism are shown.

Serhud [2]

9514 1404 393

Answer:

B. 32 square inches

Step-by-step explanation:

Each of the two triangular faces has an area of ...

A = (1/2)bh = (1//2)(6 in)(4 in) = 12 in²

The lateral surface area is the product of the prism height (0.5 in) and the perimeter of the base.

LA = (0.5 in)(5 in +5 in + 6 in) = 8 in²

So, the total area is the area of the two base and the lateral area:

SA = 2A +LA

SA = 2(12 in²) +8 in² = 32 in²

The surface area of the prism is 32 square inches.

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A line passes through the points (4,-1) and (2.3). What is the slope of the line?

deff fn [24]

Answer:

-2

Step-by-step explanation:

i got it right

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3 years ago