WILL FAN AND MEDAL Find the number halfway between 1/2 and 5/6 A.7/10 B.13/20 C.27/40 D.2/3
The half-way point between any two numbers is the average fo the numbers. To fidn the average we add the two numbers and then divide by two. \[\frac{1}{2}+\frac{5}{6}=\frac{3}{6}+\frac{5}{6}=\frac{8}{6}\]\[\frac{\frac{8}{6}}{2}=\frac{8}{6} \times \frac{1}{2}=\frac{4}{6}=\frac{2}{3}\] (D)
can you help me in some more questions?
Sure
Write the expression in simplified radical form
\[\sqrt{80}=\sqrt{5 \times 16} = \sqrt{(4 \times 4) \times 5}=4\sqrt{5}\]
Solve for x 5x^2-45=0 A.(3,-3) B.(3) C.(9) D.(9,-9)
\[5x^{2}-45=0\]\[5x^{2}=45\]\[x^{2}=9\]\[x=\sqrt{9}\]\[x=3\] Check: \[5(3^{2})-45=0\]\[5(9)-45=0\]\[45-45=0\]\[0=0\] CHECKS
Also -3 since \[(-3)^{2} = 9\] as well.
So it can just be B.(3)
Find the number halfway between:\(\dfrac{1}{2}\) and \(\dfrac{5}{6}\) \(\dfrac{1}{2} = \dfrac{3}{6}\) so What's halfway between \(\dfrac{3}{6}\) and \(\dfrac{5}{6}\)? \(\dfrac{4}{6}\) is exactly between the two fractions, and \(\dfrac{4}{6}\) reduces to \(\dfrac{2}{3}\)
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