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diamong [38]
2 years ago
6

Encuentre el angulo positivo y que es coterminal con -270

Mathematics
1 answer:
GalinKa [24]2 years ago
6 0
Ok so ummmm idk how to speck in that but well ummmmm idk
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What is the Factor 4m^3-32n^3
Klio2033 [76]

Answer:

2(47m^3-16n^3)

Step-by-step explanation:

7 0
3 years ago
What are the real or imaginary solutions of the polynomial equation x*4-25=0​
Paul [167]

Hello from MrBillDoesMath!

Answer:

Solutions: x = +\- 5i or x = +\- sqrt(5)


Discussion:

Factor x^4 - 25:

x^4 - 25     = (x^2+5) (x^2-5)                              =>  factor x^2 - 5

x^4 - 25 =   (x^2+5)(x + sqrt(5)) (x - sqrt(5))       => factor x^2 + 5

x^4 = 25  = (x +5i)(x-5i) (x + sqrt(5)) (x - sqrt(5))  

Hence the solutions are

x = +\- 5i and x = +\- sqrt(5)



Thank you,


MrB

5 0
3 years ago
Find the zeros f(x)=2x^2+9+10x
Rudiy27

Answer:

<u>−5±√7</u>

   2

hope it helped!

4 0
2 years ago
What is mZDAR in circle A?
Andreas93 [3]

Answer:

68°

Step-by-step explanation:

Since, angle subtended on the circumference of the circle is half angle subtended on the center of the circle.

\therefore m\angle DBR = \frac{1}{2} \times m\angle DAR\\\\\therefore 34\degree = \frac{1}{2} \times m\angle DAR\\\\\therefore 34\degree \times 2= m\angle DAR\\\\\therefore 68\degree = m\angle DAR\\\\\therefore m\angle DAR=68\degree

5 0
3 years ago
I don't know how to do this
Dvinal [7]
a_1=2;\ r=-\dfrac{3}{2}\\\\a_2=a_1r\to a_2=2\cdot\left(-\dfrac{3}{2}\right)=-3\\\\a_3=a_2r\to a_3=-3\cdot\left(-\dfrac{3}{2}\right)=\dfrac{9}{2}\\\\a_4=a_3r\to a_4=\dfrac{9}{2}\cdot\left(-\dfrac{3}{2}\right)=-\dfrac{27}{4}\\\\a_5=a_4r\to a_5=-\dfrac{27}{4}\cdot\left(-\dfrac{3}{2}\right)=\dfrac{81}{8}
3 0
3 years ago
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