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riadik2000 [5.3K]
3 years ago
7

The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half of its initial m

ass. Starting with 195 g of radioactive isotope how much will be left after five lives? round to the nearest gram
Mathematics
1 answer:
Dovator [93]3 years ago
4 0
1st half life: 195/2 = 97.5 g
2nd half life: 97.5/2 = 48.75 g
3rd half life: 48.7/2 = 24.375 g
4th half life: 24.375/2 = 12.1875 g
5th half life: 12.1875/2 = 6.09375 g

or you can do
195(2)^(-5) = 6.09375
** rounded to nearest gram = 6 g

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Graph AB¯¯¯¯¯¯¯¯ with endpoints A(−6, −1) and B(−5, 3) and its image after the composition.
Alona [7]

Answer:

<h3>-10⇜</h3>

Step-by-step explanation:

AB=-6×-5÷-1×3

AB= 30÷-3

<h3>AB= -10⇚</h3>

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2 years ago
Kim has exactly enough money to buy 40 oranges at 3x cents each. If the price rose to 4x cents per orange, how many oranges coul
olganol [36]

53.3

Step-by-step explanation:

Make an equation.

40y = 3x

zy=4x

Substitute y

3 0
2 years ago
How did you compute sums of dollar amounts that were not whole numbers? For example, how did you compute the sum of$5.89 and$1.4
nignag [31]

Answer:

$7.34

Step-by-step explanation:

To compute sum of dollars that are not whole numbers. Using the sum of$5.89 and$1.45 as an illustration :

$5.89 + $1.45

Taking the whole numbers first:

$5 + $1 = $6

Take the sum of the decimals :

$0.89 + $0.45 = $1.34

Sum initial whole + whole of sum of decimal

$6 + $1 = $7

Remaining decimal : $1.34 - $1 = $0.34

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4 0
3 years ago
Read 2 more answers
Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x
fredd [130]

Answer:

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

6\cdot x^{2} \geq 10 + 11\cdot x

6\cdot x^{2}-11\cdot x -10 \geq 0

6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0

Roots are found by Quadratic Formula:

r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}

r_{1} = \frac{5}{2} and r_{2} = -\frac{2}{3}

Then, the factorized form of the inequation is:

6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0

By Real Algebra, there are two condition that fulfill the inequation:

a) x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0

x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}

x \geq \frac{5}{2}

b) x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0

x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}

x\leq -\frac{2}{3}

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

3 0
3 years ago
A rocket is launched in to the air with an initial upward velocity of 375 ft/s, from ground
pshichka [43]

Answer:

h(t)=-16t^2+375t

Step-by-step explanation:

5 0
3 years ago
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