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2 years ago

I need help with this trigonometry problem.

Mathematics

1 answer:

irina1246 [14]2 years ago

4 0

The sum of all 3 angles in a triangle is 180
180-44=136
And you know that angle A is over 90 degrees so I believe it is 100 leaving angle b to be 36

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Why is it useful to identify or analyze occupational and industrial trends when considering your career path

marin [14]

It is useful. i mean because ull know how well ur going to do in the industry ur joining. if the industry is doing wee ur likely to be paid more.

5 0

3 years ago

Read 2 more answers

Quiz 3

Colt1911 [192]

Kevin would be 24 because 4 times 6 equal to 24

3 0

3 years ago

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The sum of a number times 7 and 21 is at least 29. Write as an inequality

PtichkaEL [24]

The inequality is 7x + 21 $\leq 29$

Step-by-step explanation:

DATA:

Let the number be x

Number times 7 = 7*x

Sum of a number times 7 and 21 = 7x + 21

Is at least 29

INEQUALITY:

7x + 21 $\leq 29$

SOLVE:

7x $\leq 29-21$

7x $\leq 8$

x $\leq \frac{8}{7}$

Therefore, the inequality is 7x + 21 $\leq 29$

Keyword: Inequality

Learn more about inequality at

#LearnwithBrainly

6 0

3 years ago

Solve for y if: x=2 y=5x+3

nika2105 [10]

Substitute in x = 2:

y = 5(2) + 3

y = 10 + 3

y = 13

4 0

3 years ago

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Alexander Litvinenko was poisoned with 10 micrograms of the radioactive substance Polonium-210. Since radioactive decay follows

koban [17]

Answer:

The amount of Polonium-210 left in his body after 72 days is 6.937 μg.

Step-by-step explanation:

The decay rate of Polonium-210 is the following:

$N(t) = N_{0}e^{-\lambda t}$ (1)

Where:

N(t) is the quantity of Po-210 at time t =?

N₀ is the initial quantity of Po-210 = 10 μg

λ is the decay constant

t is the time = 72 d

The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.

First, we need to find the decay constant:

$\lambda = \frac{ln(2)}{t_{1/2}}$ (2)

Where t(1/2) is the half-life of Po-210 = 138.376 days

By entering equation (2) into (1) we have:

$N(t) = N_{0}e^{-\frac{ln(2)}{t_{1/2}}*t}} = 10* \frac{99.498}{100}*e^{-\frac{ln(2)}{138.376}*72} = 6.937 \mu g$

Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.

I hope it helps you!

8 0

2 years ago