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alexgriva [62]
3 years ago
13

4j - 24- 6j = 120 what's the solution? solution=

Mathematics
1 answer:
maw [93]3 years ago
4 0
4j - 24- 6j = 120\\
4j-6j=120+24\\-2j=144\\
j=-72
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You select a letter randomly from a bag containing the letters s, p, i, n, n, e, and r. find the probability of selecting an s.
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1/7

Step-by-step explanation:

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What are the discontinuity and zero of the function f(x) = quantity x squared plus 6 x plus 8 end quantity over quantity x plus
Tcecarenko [31]

Answer:

Discontinuity at (-4,-2), zero at (-2,0).

Step-by-step explanation:

We are given that a function

f(x)=\frac{x^2+6x+8}{x+4}

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f(-4)=-4+2=-2

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Zero: The zero of the function is that number when substitute it in the given function then the function becomes zero.

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In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!
g100num [7]

the real solutions for the equation x^{3}=-64 are -

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Step-by-step explanation:

   x^{3} = - 64

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   We can write 64 as  4^{3}

  x^{3} + 4^{3} = 0

  using the identity  ( x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} ) )

we get,

  = (x+4) (x^{2} -x*4+4^{2} )

  = (x+4)(x^{2} -4x+16)    ....................(1)

 solving the quadratic equation  ,

   x^{2} -4x+16 =0

solutions of this quadratic equation can be obtained by

   x=-b +- \sqrt{b^{2}-4ac } /2a

let use y for factors

x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16}  / 2*x^{2}

x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}

x=4+-\sqrt{16-64}/2

x=4+-4\sqrt{3i} /2

<u />x=2+-2\sqrt{3i}    ..................(2)

from the equation 1 we have,

x-4=0

which gives solution x=4

and from equation 2 we got  x=2+-2\sqrt{3i}

so the real solutions for the equation x^{3}=-64 are -

x=4,2+-2\sqrt{3i}

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