All categories

3 years ago

Ashley Sigler gave 2/5 of her bonus check of $255 to a missionary family. How much

Mathematics

1 answer:

STALIN [3.7K]3 years ago

8 0

Answer:

102

Step-by-step explanation:

i think it is right

You might be interested in

Jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

skad [1K]

Answer:

Step-by-step explanation:

2+2=2 and then 1 come now you have 1 2 and 1 1

5 0

2 years ago

I NEED HELP PLEASE ! :)

dedylja [7]

We can write a proportion to resemble the problem;

AE/ED = AB/BC

AE = 9

ED = 6

AB = x

BC = 10

Substitute with the given values.

9/6 = x/10

9/6 * 10 = x/10 * 10

90/6 = x

15 = x

Therefore, the answer is 15.

Best of Luck!

7 0

3 years ago

Read 2 more answers

What is 8 1/8 divided by 5/6

mart [117]

Answer:

9.75

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Chile we do be small brain around here ✌

umka2103 [35]

Answer:

chile I think it's A

Step-by-step explanation:

anyway, if it goes through the same point as the the first one, it becomes a solution and depending on the equation everything either to the left or right, above or below, becomes a solution as well. don't stress bruv, we all got small brain

3 0

2 years ago

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

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

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

$(1-r)^{5.3} = 0.5$

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

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

$1 - r = 0.8774$

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

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