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Pani-rosa [81]
3 years ago
15

Describe the solutions of 5

Mathematics
1 answer:
djyliett [7]3 years ago
6 0

Answer:

A value, or values, we can put in place of a variable (such as x) that makes the equation true.

Example: x + 2 = 7

When we put 5 in place of x we get: 5 + 2 = 7

5 + 2 = 7 is true, so x = 5 is a solution

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What is [b-5] over 8=5
julia-pushkina [17]

Answer:

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8 0
2 years ago
Find all the values of x in the set of complex numbers that satisfy the following equation:
Fynjy0 [20]
Compute all the component integrals first:

I_1=\displaystyle\int_0^{\pi/4}2\sec x\,\mathrm dx=2\ln(\sqrt2+1)
I_2=\displaystyle\int_0^2\ln x\,\mathrm dx=2(\ln2-1)
I_3=\displaystyle\lim_{a\to\infty}\int_{-a}^a\frac{\mathrm dx}{x^2+1}=\pi

Now,

\sqrt2\approx1.4\implies \sqrt2+1\approx2.4
\implies \left\lceil I_1\right\rceil=2

e
\implies\left\lfloor I_2\right\rfloor=-1

\pi\approx3.14\implies\left\lceil I_3\right\rceil=4

So the given equation reduces to

\displaystyle\sum_{k=-1}^2\frac{\mathrm d}{\mathrm dx}x^{k+2}=1-4!
\dfrac{\mathrm dx}{\mathrm dx}+\dfrac{\mathrm dx^2}{\mathrm dx}+\dfrac{\mathrm dx^3}{\mathrm dx}+\dfrac{\mathrm dx^4}{\mathrm dx}=-23
4x^3+3x^2+2x+24=0

a fairly standard cubic. Incidentally, when x=-2, the LHS reduces to 0, so x+2 is a factor of the cubic. You can find the remaining two solutions easily with the quadratic formula.
5 0
3 years ago
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