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3 years ago

10

A set of three numbers with the greatest common factor of 13

1 answer:

BigorU [14]3 years ago

3 0

78, 130, and 91. You're welcome :D

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How do you solve this?

Naddika [18.5K]

$\dfrac{3+4i}{2+i}=\dfrac{(3+4i)(2-i)}{(2+i)(2-i)^{(*)}}=\dfrac{3\cdot2-3\cdot i+4i\cdot2-4i\cdot i}{2^2-i^2^{(**)}}\\\\=\dfrac{6-3i+8i+4}{4-(-1)}}=\dfrac{10+5i}{4+1}=\dfrac{10+5i}{5}=\dfrac{10}{5}+\dfrac{5i}{5}=\boxed{2+i}\to \boxed{a=2}\\\\used:\\(*)\ (a+b)(a-b)=a^2-b^2\\(**)\ i^2=-1$

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vladimir2022 [97]

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4/10 or 40%

Step-by-step explanation:

there are 4 balls with the number 3. 4 balls out of 10. 4/10.

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