Peter’s dog eats 0.3 kg of dog food each day how much dog food will Peter’s dog eat in eight days

What equivalent fractions for 2 3/5 and 1 1/3 using a common denominator

inysia [295]

2 9/15 and 1 5/15 are your answers

8 0

3 years ago

Read 2 more answers

Need answer thanks! only answer if you know please!

Vaselesa [24]

Answer:

a) $k \approx 0.0602\,\frac{1}{min}$ ( $k\approx 6.02\,\frac{\%}{min}$ ), b) $k \approx 0.0648\,\frac{1}{min}$ ( $k \approx 6.48\,\frac{\%}{min}$ ), c) $t_{1/2} \approx 60.274\,days$ , d) $t_{1/2}\approx 4.305\,min$ , e) $k \approx 0.3\,\frac{1}{min}$ ( $k\approx 30\,\frac{\%}{min}$ ), f) $k \approx 0.0120\,\frac{1}{min}$ ( $k \approx 1.20\,\frac{\%}{min}$ ), g) $k \approx 2.876\times 10^{-5}\,\frac{1}{yr}$ ( $k \approx 2.876\times 10^{-3}\,\frac{\%}{yr}$ )

Step-by-step explanation:

All radioactive isotopes decay exponentially, the mass of the isotope as a function of time ( $m(t)$ ), measured in grams, is defined by:

$m(t) = m_{o}\cdot e^{-k\cdot t}$ (1)

Where:

$m_{o}$ - Initial mass of the isotope, measured in grams.

$k$ - Decay rate, measured in $\frac{1}{min}$ or $\frac{1}{day}$ or $\frac{1}{yr}$ .

$t$ - Time, measured in minutes, days or years.

We define the decay rate in terms of half-life by using the following expression:

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

Where $t_{1/2}$ is the half-life of the isotope, measured in minutes or years.

Now we proceed to determine the missing values:

a) Polonium-200 ( $t_{1/2} = 11.5\,min$ )

$k = \frac{\ln 2}{11.5\,min}$

$k \approx 0.0602\,\frac{1}{min}$

b) Lead-194 ( $t_{1/2} = 10.7\,min$ )

$k = \frac{\ln 2}{10.7\,min}$

$k \approx 0.0648\,\frac{1}{min}$

c) Iodine-125 ( $k = 0.0115\,\frac{1}{day}$ )

$t_{1/2} = \frac{\ln 2}{k}$

$t_{1/2} = \frac{\ln 2}{0.0115\,\frac{1}{day} }$

$t_{1/2} \approx 60.274\,days$

d) Kryption-75 ( $k = 0.161\,\frac{1}{min}$ )

$t_{1/2} = \frac{\ln 2}{k}$

$t_{1/2} = \frac{\ln 2}{0.161\,\frac{1}{min} }$

$t_{1/2}\approx 4.305\,min$

e) Strontium-79 ( $t_{1/2} = 2.3\,min$ )

$k = \frac{\ln 2}{2.3\,min}$

$k \approx 0.3\,\frac{1}{min}$

f) Uranium-229 ( $t_{1/2} = 58\,min$ )

$k = \frac{\ln 2}{58\,min}$

$k \approx 0.0120\,\frac{1}{min}$

g) Plutonium-239 ( $t_{1/2} = 24100\,yr$ )

$k = \frac{\ln 2}{24100\,yr}$

$k \approx 2.876\times 10^{-5}\,\frac{1}{yr}$

6 0

3 years ago

What are the solutions of 3(x – 4)(2x - 3) = 0? Check all that apply. -4 -3 win w bo 3 m/N 3 4

Fudgin [204]

Answer:

Option (4) and Option (6)

Step-by-step explanation:

By the zero product property,

If (x - a)(x - b) = 0

Then, (x - a) = 0 ⇒ x = a

(x - b) = 0 ⇒ x = b

By following the zero product property,

From the given equation,

3(x - 4)(2x - 3) = 0

x - 4 = 0 ⇒ x = 4

2x - 3 = 0 ⇒ x = $\frac{3}{2}$

Therefore, options (4) and (6) will be the correct options.

7 0

3 years ago

Read 2 more answers

Which of the following expression could be substituted for the height of the triangle in order to derive the formula Area=1/2ab

noname [10]

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8 0

3 years ago

A - 15 = 22 what the value of a

svetlana [45]

a - 15 = 22

a = 22 + 15

a = 37

The value of a is 37

7 0

3 years ago