You should get this answer because the simpliest form of a radical is that you shouldn't have any radical in the denominator. In order to get rid of the radical sign in the denominator, you have to multiply the numerator and the denominator with the $\sqrt{5}$ .

Look at the image below for further explanation

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6 0

2 years ago

Intersection of which lines should be taken as the center of the circumscribed circle of a triangle?

mel-nik [20]

The perpendicular bisectors.

6 0

3 years ago

Help me in question 7

andriy [413]

I think it’s either b or c

8 0

3 years ago

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If 150 grams of a radioactive isotope are present at 2:00 PM and 10 grams remain at 6:00 PM (the same day), what is the half-lif

pogonyaev

Initial amount, $A_o=150\ g$ .

Final amount, $A =10\ g$ .

Time taken, t = 6:00 - 2:00 = 4 hour.

We know,

$A=A_o(\dfrac{1}{2})^{\dfrac{t}{h}}\\\\10 = 150 \times \dfrac{1}{2}^{\dfrac{4}{h}}\\\\2^{\dfrac{4}{h}}=15\\\\\dfrac{4}{h}= log_215\\\\h = \dfrac{4}{log_215}\\\\h = 1.024 \ hours$

Therefore, the half-life of the isotope is 1.024 hours.

Hence, this is the required solution.

8 0

2 years ago