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Goshia [24]
3 years ago
5

Find the circumference of the circle to the nearest tenth.

Mathematics
1 answer:
Y_Kistochka [10]3 years ago
7 0
C = 2(PI)r multiply 2 x 3.14 x (whatever the radius is!
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Write (3i - 2/3) - (6i-5) as a complex number in standard form. I have no idea what im doing to be honest -_-. Give me pointers
allochka39001 [22]
\bf \left( 3i-\cfrac{2}{3} \right)-(6i-5)\implies 3i-\cfrac{2}{3}-6i+5
\\\\\\
-6i+5-\cfrac{2}{3}\implies 
\begin{array}{llll}
\cfrac{13}{3}&-&6i\\
\uparrow &&\uparrow \\
a&-&bi
\end{array}

standard form for a complex expression, I assume it means the a + bi form, nothing else

5 0
3 years ago
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Help Pls! It's for my Math 4 class! Distance on the complex plane!
Serhud [2]
Picture is not so clear it’s blur

8 0
3 years ago
Y = -2x - 3, provide the x-y coordinates of the y-intercept and the x-intercept. show your work.
PilotLPTM [1.2K]
Y intercept:
Y=-2(0)-3
Y=-3
(0,-3)

X-intercept
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3=-2X
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(-3/2,0)
5 0
3 years ago
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How can I find the complex roots of 125x^3+343. Please explain your steps
8_murik_8 [283]
First we assume that this is equal to zero

so what we notice is this is a sum of 2 perfect cube

recall that
a^3+b^3=(a+b)(a^2-ab+b^2)

125x^3+343=
(5x)³+(7)³=(5x+7)(25x²-35x+49)

now solve the part in second parenthasees because we can see that the first parenthasees (5x+7) is going to have a real root

remember quadratic formula

for ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
a=25
b=-35
c=49

x=\frac{-(-35)+/- \sqrt{(-35)^2-4(25)(49)} }{2(25)}
x=\frac{35+/- \sqrt{1225-4900} }{50}
x=\frac{35+/- \sqrt{-3675} }{50}
x=\frac{35+/- (\sqrt{3675})( \sqrt{-1}) }{50}
x=\frac{35+/- (\sqrt{3675})(i) }{50}
x=\frac{35+/- 35i\sqrt{3} }{50}
x=\frac{7+/- 7i\sqrt{3} }{10}

the comlex roots are

x=\frac{7+ 7i\sqrt{3} }{10} and \frac{7- 7i\sqrt{3} }{10}




3 0
3 years ago
2 please answer this question
Fittoniya [83]
B. End of the reconstruction. 
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4 years ago
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