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3 years ago

Please find x for numbers 1 & 2. Greatly appreciated:)

Mathematics

1 answer:

gavmur [86]3 years ago

7 0

Answer:

1. 14.5

2. 16

Step-by-step explanation:

1. a^2 + b^2 =c^2

20^2 + 21^2 = c^2

400 + 441 = c^2

841 = c^2

$\sqrt{841}$ = $\sqrt{c}$

29 = c

29/2 = x

14.5 =x

2. c^2-b^2 = a^2

34^2 - 30^2 = a^2

1156 - 900 = a^2

256 = a^2

$\sqrt{ 256 }$ = $\sqrt{a}$

16 = a

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Oh lord help :,) please and thank you

ziro4ka [17]

Answer:

The answer is 19.10 $

Step-by-step explanation:

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3 years ago

Read 2 more answers

Assume you have such a watch. if a minimum of 18.0% of the original tritium is needed to read the dial in dark places, for how m

laila [671]

Given that the half life of 3H is 12.3 years.

The amount of substance left of a radioactive substance with half life of $t_{ \frac{1}{2} }$ after t years is given by

$N(t)=N_0\left(\frac{1}{2} \right)^{ \frac{t}{t_{\frac{1}{2}} }$

Therefore, the number of years it will take for 18% of the original tritinum to remain is given by

$18 \% = 100 \% \left(\frac{1}{2} \right)^{ \frac{t}{12.3}} \\ \\ \Rightarrow\left(\frac{1}{2} \right)^{ \frac{t}{12.3}} =0.18 \\ \\ \Rightarrow\frac{t}{12.3}\ln\left(\frac{1}{2} \right)=\ln0.18 \\ \\ \Rightarrow\frac{t}{12.3}= \frac{\ln0.18}{\ln\left(\frac{1}{2} \right)} = \frac{-1.715}{-0.6931} =2.474\\ \\ \Rightarrow t=2.474(12.3)=30.4$

Therefore, the number of <span>years that the time could be read at night is 30.4 years.</span>

5 0

3 years ago

Find the y-intercept of the line -4× - 2y =16

Oksi-84 [34.3K]

Answer:

The y intercept is (0,-8)

Step-by-step explanation:

To find the y intercept, set x = 0 and solve for y

-4x -2y = 16

-4(0) -2y =16

-2y = 16

Divide by -2