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julsineya [31]
3 years ago
9

Find all the real square roots of 400. 20 -20 and 20 -40 -40 and 40

Mathematics
1 answer:
Scrat [10]3 years ago
4 0

Answer:

  The square roots of 400 are 20 and -20.

Step-by-step explanation:

You can try the offered answers to see if they square to give 400:

  (-40)² = 1600 . . . . -40 is not a square root of 400

  (-20)² = 400 . . . . -20 is a square root of 400

  20² = 400 . . . . . 20 is a square root of 400

  40² = 1600 . . . . 40 is not a square root of 400

_____

The second degree equation ...

  x² = 400

will have exactly two solutions, as required by the fundamental theorem of algebra. Those solutions are both given above. There are no other square roots of 400, real or complex.

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Answer:

s = 3

Step-by-step explanation:

we know the rectangles are similar

2 × __ = 16

s × __ = 24

so,

16÷2 = 8

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then,

24 ÷ 8 = s

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hoped i helped :)

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Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separ
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Answer:

\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}

Step-by-step explanation:

Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}

\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16

\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i

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3 years ago
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