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Sergeu [11.5K]
2 years ago
10

What iis the square root of 158

Mathematics
2 answers:
IrinaVladis [17]2 years ago
6 0
158 is not a perfect square
About 12.5698051
never [62]2 years ago
3 0

Answer:

The answer is 12.56, which can be rounded to 12.7 or 13. :)

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Solve sec2theta + tantheta - 3 =0 for 0
Serhud [2]

Answer:

Ф = \frac{\pi }{4} ,\frac{5\pi}{4}

Step-by-step explanation:

It is a bit difficult to input the work here, so I uploaded an image

  • First we can use the trig identities to change sec²(Ф) to tan²(Ф) + 1
  • Then we can combine like terms
  • Then we can factor this as a polynomial function
  • Then we can set each term equal to zero and solve for Ф
  • The first term tan(Ф) - 2 = 0 has no solution because tan(Ф) ≠ -2 anywhere
  • The second term tan(Ф) - 1 = 0 has two solutions of \frac{\pi}{4} and \frac{5\pi}{4} so these are the solutions to the problem

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Consider the following division of polynomials.
Bond [772]

x^4=x^2\cdot x^2. Multiplying the denominator by x^2 gives

x^2(x^2+2x+8)=x^4+2x^3+8x^2

Subtracting this from the numerator gives a remainder of

(x^4+x^3+7x^2-6x+8)-(x^4+2x^3+8x^2)=-x^3-x^2-6x+8

-x^3=-x\cdot x^2. Multiplying the denominator by -x gives

-x(x^2+2x+8)=-x^3-2x^2-8x

and subtracting this from the previous remainder gives a new remainder of

(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8

This last remainder is exactly the same as the denominator, so x^2+2x+8 divides through it exactly and leaves us with 1.

What we showed here is that

\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}

=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}

=x^2-x+1

and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)

=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)

=x^4+x^3+7x^2-6x+8

which matches the original numerator.

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Step-by-step explanation:

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