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oksian1 [2.3K]
3 years ago
5

Carter was given a box of assorted chocolates for his birthday. Each night, Carter treated himself to some chocolates. Carter at

e 5 chocolates each night and there were originally 30 chocolates in the box. Write an equation for C,C, in terms of t,t, representing the number of chocolates remaining in the box tt days after Carter's birthday.
Mathematics
1 answer:
yanalaym [24]3 years ago
3 0
C(t) = 30 - 5t ur very much welcome
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The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

N = N_0 * (\frac{1}{2})^{t/p

N_0 is the starting amount of parent element.

N is the end amount of parent element

t is the time elapsed

p is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

N(t) = 74g * (\frac{1}{2})^{t/2.8days

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2}  =  37g

c.

Now we just gotta do some algebra. Use the original function but this time, replace N(t) with 10g and solve algebraically.

10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}

Take the log of both sides.

log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})

Use the exponent rule for log laws that, log(b^x) = x*log(b)

log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})

\frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = \frac{t}{2.8days}

2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = t

slap that in your calculator and you get

t = 8.1 days

7 0
2 years ago
There are three monomials such that the greatest common factor of the first and second monomials is 2xy, and the greatest common
AlekseyPX

<span>let: m1 = 2xy a m2 = 2xy b m3 = 2x2y c since 2 and 3 have a common factor let b=2n m1 = 2xy a m2 = 2xy 2 n m3 = 2x2y c or simply m1 = 2xy a m2 = 2xy 2n m3 = 2xy 2c given that anc have no common factors</span>
6 0
3 years ago
During a checkers game, there are 16 pieces left. The ratio of black to red is 3 : 5. How many black pieces are on the board? Ex
kotegsom [21]

Answer:

6 black pieces

10 red pieces

Step-by-step explanation:

The way to interpret ratios is like so:

Adding up both parts of the ratio gives 8, i.e. 3 + 5

So for every 8 pieces, 3 are black and 5 are red;

There are 16 pieces, which is 2 × 8;

So, there are (2 × 3) black pieces and (2 × 5) red pieces, i.e. 6 black and 10 red

So, to solve:

Add all parts of the ratio (3 + 5);

Divide the number of pieces (16) by this sum of ratio parts (8);

Multiply the value from the division (2) by each part of the ratio

Tip:

It is important to take note of the order in the ratio, so the ratio of black to red means the number that comes first in the ratio represents parts black and the one that comes second represents parts red;

If the wording was the ratio of red to black, the number that comes first would represent parts red and the second would represent parts black

3 0
3 years ago
Help me with this math question
nasty-shy [4]

Answer:

\huge \boxed{K=20}\checkmark

B. 20

Step-by-step explanation:

<em>First, you divide by 5 from both sides of equation.</em>

<em>\displaystyle \frac{5k}{5}=\frac{100}{5}</em>

<em>Simplify, to find the answer.</em>

<em>\displaystyle 100\div5=20</em>

<em>\huge \boxed{K=20}, which is our answer.</em>

6 0
3 years ago
Read 2 more answers
Which of the following equations is an example of inverse variation between
AURORKA [14]

Answer:

B

Step-by-step explanation:

Any inversely proportional equation is of the form y=k/x.

5 0
3 years ago
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