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2 years ago

1 year = how many seconds? Last question in my h.w and idk how to do it:/

Mathematics

1 answer:

stiv31 [10]2 years ago

8 0

There is about 31 million seconds in 1 year

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What is the slope intercept form of 12x-5y=-44

Lilit [14]

Answer:

y = $\frac{12}{5}$ x + $\frac{44}{5}$

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

rearrange 12x - 5y = - 44 into this form

subtract 12x from both sides

- 5y = - 12x - 44 ( divide all terms by - 5 )

y = $\frac{12}{5}$ x + $\frac{44}{5}$ ← in slope- intercept form

4 0

2 years ago

What is 2.26 (26 repeating)as a fraction

irina [24]

Well 2.26 as a fraction is
$2 \times \frac{13}{50}$
^Whole Fraction^

but as an Improper Fraction I is
$\frac{113}{50}$

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2 years ago

The children's section is rectangular and has three of its comers at (-10,10). (-10,4), and (-3,10). What are the coordinates

oee [108]

Answer:

(-3, 4)

Step-by-step explanation:

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2 years ago

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IrinaK [193]

Could you please give me more of a description?

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2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

2 years ago