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nikdorinn [45]
3 years ago
9

If 12g of a radioactive substance are present initially and 8yr later only 6g remain, how much if the substance will be present

after 13 years?
Mathematics
1 answer:
Brums [2.3K]3 years ago
5 0

The weight of substance remaining after 13 years is 2.25 g.

Given:

Present weight of substance = 12 g

After 8 years weight remains = 6 gm

Now, weight loss in 8 year will be,

12-6=6 gm

So, weight loss in 1 year will be,

\dfrac{6}{8} =\dfrac{3}{4}

Total weight loss in 13 year will be,

\dfrac{3}{4}\times13=9.75

The weight of substance remaining after 13 years will be,

12- 9.75 = 2.25 \;\rm g

Therefore, the weight of substance remaining after 13 years is 2.25 g.

To know more about radioactive decay, refer to the link:

brainly.com/question/12116123

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Answer:

Pretty sure it's 20.24.

Step-by-step explanation:

Since you know the side adjacent to the reference angle and you want to find the side opposite, you're going to need to use tangent, which is opp/adj.

tan(39) = \frac{x}{25}

Find the tangent of 39...

0.8097 is the approximate value. So...

0.8097 = \frac{x}{25}

Now, switch the sides to make it easier.

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Multiply both sides by 25.

x = 20.24

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Answer:

Step-by-step explanation:

LHS =\dfrac{1}{Sin^{2} \ A }-\dfrac{1}{Tan^{2} \ A }\\\\\\ = \dfrac{1}{sin^{2} \ A}- \dfrac{1}{\dfrac{Sin^{2} \ A}{Cos^{2} \ A}}\\\\\\= \dfrac{1}{sin^{2} \ A } - \dfrac{Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{1-Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{Sin^{2} \ A}{Sin^{2} \ A}\\\\\\= 1 = \ RHS

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Consider triangle ABC where AB=X+5, BC=X-2, area is 30cm squared. Find X by solving X^2+3X-70=0, and find the perimeter.
klasskru [66]

Answer:

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Step-by-step explanation:

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       x+10=0

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Area of ABC = 30cm2

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<u />

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