<img src="https://tex.z-dn.net/?f=%5Csqrt%7B180%7D%3Da%5Csqrt%7Bb%7D" id="TexFormula1" title="\sqrt{180}=a\sqrt{b}" alt="\sqrt{1

$4\dfrac{2}{3}\div4\dfrac{1}{5}=\dfrac{14}{3}\div\dfrac{21}{5}=\dfrac{14\!\!\!\!\!\diagup^2}{3}\cdot\dfrac{5}{21\!\!\!\!\!\diagup_3}=\dfrac{2\cdot5}{3\cdot3}=\dfrac{10}{9}=1\dfrac{1}{9}$

8 0

3 years ago

EASY QUESTION! <br><br>What is the value of t?<br><br>-5 = 2t - 11

myrzilka [38]

$-5 = 2t - 11\\\\2t=6\\\\t=3$

4 0

4 years ago

Read 2 more answers

How many, and what type of solutions, does3x2−x−5=0 have?

siniylev [52]

Answer:

2 irrational solutions

Step-by-step explanation:

using the quadratic formula:

$\frac{ -b \:± \sqrt{b {}^{2} - 4ac } }{2a}$

3x²−x−5=0

where a = 3, b = -1 and c = -5, plug in those values in the quadratic formula

$\frac{ - ( - 1)±\sqrt{(1) {}^{2} - 4(3)( - 5) } }{2(3)}$

you get two solutions:

$\frac{1 + \sqrt{61} }{6} \: and \: \frac{1 - \sqrt{61} }{6}$

both can be written as:

$\frac{1± \sqrt{61} }{6}$

They are irrational as they cannot be written in the form p/q where p and q are whole numbers (integers)

8 0

3 years ago