A system of two linear equations and three unknowns can be solved by:

Mathematics

1 answer:

irakobra [83]3 years ago

3 0

These cannot be solved.- 3 unknowns require 3 equations

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Step-by-step explanation:

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What is the next term in this geometric sequence<br>5/6,5/9,10/27,20/81

Ghella [55]

$a_1=\frac{5}{6};\ a_2=\frac{5}{9};\ a_3=\frac{10}{27};\ a_4=\frac{20}{81}\\\\\\q=a_2:a_1\\\\q=\frac{5}{9}:\frac{5}{6}=\frac{5}{9}\cdot\frac{6}{5}=\frac{2}{3}\\\\\\a_5=a_4\cdot q\\\\a_5=\frac{20}{81}\cdot\frac{2}{3}=\frac{40}{243}\to answer$

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3 years ago

Read 2 more answers

HELP ASAP PLEASE & SHOW UR WORK!! THANK YOU <333

alekssr [168]

Answer:

$\bold{x_1=-\dfrac{3+\sqrt{41}}{2}\approx-4.7\ ,\quad x_2=-\dfrac{3-\sqrt{41}}{2}\approx1.7}$

Step-by-step explanation:

$-2x^2-3x+4=0\\\\a=-2\,,\ b=-3\,,\ c=4\\\\\\x=\dfrac{-b-\sqrt{b^2\pm4ac}}{2a}=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(-2)(4)}}{2(-1)}=\dfrac{3\pm\sqrt{9+32}}{-2}\\\\\\x_1=-\dfrac{3+\sqrt{41}}{2}\approx-4.7\ ,\qquad x_2=-\dfrac{3-\sqrt{41}}{2}\approx1.7$