Answer:
b
Step-by-step explanation:
sqrt (3/10)
sqrt(3)/sqrt(10)
no sqrt in the denominator, so multiply by sqrt(10)/sqrt(1)
sqrt(3) sqrt(10)
-------------------
sqrt(10) sqrt(10)
sqrt(30)/10
Choice D
Answer:
√45
Step-by-step explanation:
We know that according to the Pythagorean theory:
a²+b²=c²
I am pretty sure that a is 3 and b is 6. C remains unknown.
3 squared is nine and 6 squared is thirty six. 36+9 is 45.
Since the question says you should leave the answer in radical form, the answer is
c²=45
√c²=√45
This book will really help in Geometry and you can get it anywhere.
I hope this helps!
~°LampteyJ
Almost 60 years after taking my last math class, I discovered
how to turn a repeating decimal into its equivalent fraction.
-- Take the set of digits that repeats, no matter how many digits there are.
Write them as the numerator of a fraction.
-- For the denominator, write the same number of '9's.
-- If it's possible and you feel like it, simplify (reduce) the fraction.
You have 0.023 with the 23 (two digits) repeating forever.
-- Write the 23 (two digits) as the numerator.
-- Write 99 (two digits) as the denominator.
Now you have the fraction 23 / 99 . It can't be reduced.
You actually had 0.0... before the repeating part. That just
means that the fraction has been divided by 10. The real
equivalent fraction is 23 / 990 . (Which also can't be reduced.)
After so many years not knowing how to do it, this still blows my mind.
Just now, I punched " 23/990 "into my calculator, and stared in wonder
and amazement when " 0.023232323 " came up.
Answer:
ΔGJH ≅ ΔEKF
HL: GH and EF
SAS: FK and JH (or GH and EF)
ASA: ∠JGH and ∠FEK (or ∠EFK and ∠JHG)
ΔGFJ ≅ ΔEKH
SSS: KH and FJ
SAS: ∠KEH and ∠FGJ
Step-by-step explanation:
List whatever angles/sides need to be congruent for the two triangles to be congruent.
Prove ΔGJH ≅ ΔEKF using....
- HL (Hypotenuse + Leg)
We already have two legs that are congruent (EK and GJ), so we just need the hypotenuses (GH and EF) to be equal.
- SAS (Side + Angle + Side)
1 pair of sides (EK and JG) are equal, and m∠EKF = m∠GJH. So we need one more side. You can either use FK and JH or GH and EF.
- ASA (Angle + Side + Angle)
1 pair of angles (∠EKF and ∠GJH) are already given as equal, and 1 pair of sides (EK and GJ) are equal. We just need one more pair of angles. So either ∠JGH and ∠FEK or ∠EFK and ∠JHG.
Prove ΔGFJ ≅ ΔEKH using...
- SSS (Side + Side + Side)
Two pairs of sides (EK + GJ and EH + FG) are equal, so KH and FJ need to be equal.
- SAS (Side + Angle + Side)
FG + EH and KE + GJ are equal. We need to use the angle in between them to use SAS, so ∠KEH and ∠FGJ need to be equal.
