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kotegsom [21]
3 years ago
13

Please solve and graph each inequality x-7<-12

Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0

For this case we must resolve the following inequality:

x-7

Adding 7 to both sides of the inequality:

x

Different signs are subtracted and the major sign is placed.

x

Thus, the solution is given by all the values of "x" less than -5.

The solution set is: (-∞, - 5)

Answer:

x

See attached image

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Let f be a functions of degree 4 whose coefficients are real numbers: two of its zeros are - 3 and 4 - i. Explain why one of the
kvasek [131]

Answer:

Step-by-step explanation:

We have the following theorem, if f is a polynomial with real coefficients, we can factor it completely in factors of the degree at most 2.

Consider first a polynomial of degree two, hence it is a polynomial of the form ax^2+bx+c. The cuadratic formula tells us that the solutions are of the form

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x = \frac{-b \pm i \sqrt[]{4ac-b^2}}{2a}. NOte that this means that if we have a complex number of the form a+bi that is a solution, then the number a-bi (who is called the complex conjugate) is also a solution.

Recall that when we have a polynomial f(x) whose a zero is the number c, then we can factor f as follows f(x) = (x-c) * p(x) where p(x) is another polynomial of lesser degree .

So far, we know that -3 and 4-i are zeros of the function f. Note that we are missing two zeros. But, since complex numbers are zeros of polynomial only by pairs (that is the number and its conjugate are zeros), then, we must have that one of the missing 2 zeros is a real number. We have 4-i as a zero, then, its complex conjugate must be also a zero, i.e 4+i is a zero.

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Solve the initial value problem yy′+x=x2+y2‾‾‾‾‾‾‾√ with y(1)=8‾√. To solve this, we should use the substitution u= x^2+y^2 help
leva [86]

Looks like the equation is

yy'+x=\sqrt{x^2+y^2}

Substitute u(x)=x^2+y(x)^2, so that u'=2x+2yy'. Then the equation is the same as

\dfrac{u'-2x}2+x=\sqrt u\implies u'=2\sqrt u\implies\dfrac{\mathrm du}{2\sqrt u}=\mathrm dx

Integrate both sides to get

\sqrt u=C\implies\sqrt{x^2+y^2}=C

Given that y(1)=\sqrt8, we have

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so the solution is

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