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devlian [24]
4 years ago
7

Help me please with 3-111

Mathematics
1 answer:
LenKa [72]4 years ago
4 0

Last Question

Because this may not be the question you want, I'll omit the first step which should be done with latex. The key to all of these except C is to invert the second fraction (turn the second fraction upside down) and multiply.

A

\dfrac{6}{5} * \dfrac{-2}{3}

=(6*-2) / (5*3) = - 12/15 = -4/5

B

\dfrac{7}{4} *\dfrac{5}{3}

= (7*5) / (4*3) = 35 / 12 = 2 11/12

C

Impossible to do 4/0 cannot be done.

D

\dfrac{-2}{3} /\dfrac{-5}{4}=\dfrac{-2}{3} *\dfrac{-4}{5}

= (-2*-4)/(3*5) =

8/15


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Question 3 (1 point) Co-60 has a half life of 5.3 years. If a pellet that has been in storage for 26.5 years contains 14.5g of C
Zarrin [17]

Answer:

464 grams.

Step-by-step explanation:

Amount of substance:

The amount of substance after t years is given by an equation in the following format:

A(t) = A(0)(1-r)^t

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Co-60 has a half life of 5.3 years.

This means that:

A(5.3) = 0.5A(0)

We use this to find r. So

A(t) = A(0)(1-r)^t

0.5A(0) = A(0)(1-r)^{5.3}

(1-r)^{5.3} = 0.5

\sqrt[5.3]{(1-r)^{5.3}} = \sqrt[5.3]{0.5}

1 - r = 0.5^{\frac{1}{5.3}}

1 - r = 0.8774

So

A(t) = A(0)(0.8774)^{t}

If a pellet that has been in storage for 26.5 years contains 14.5g of Co-60, how much of this radioisotope was present when the pellet was put in storage?

We have that A(26.5) = 14.5, and use this to find A(0). So

A(t) = A(0)(0.8774)^{t}

14.5 = A(0)(0.8774)^{26.5}

A(0) = \frac{14.5}{(0.8774)^{26.5}}

[tex]A(0) = 464[tex]

464 grams.

3 0
3 years ago
Please someone help me
polet [3.4K]
1)
2(3x-5)
Use the distributive property.
2(3x)+2(-5)
6x-10 
That is not equal to 6x-8.

2)
2-2+5x
Simplify the like terms, 2 and -2
Add them together, 0
5x=5x the two expressions are equal

3)
2x+8 because there are x stickers in a pack
2(4+x)=2(4)+2(x)=8+2x=2x+8
Yes they are the same expressions.


6 0
3 years ago
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