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3 years ago

Who makes faces all day long DUE AT 8:30 PLZZZZZZ

Mathematics

1 answer:

sveta [45]3 years ago

6 0

Answer:

The answer to this joke is a clockmaker.

Step-by-step explanation:

The math is just finding volume. You are asking for just the answer correct?

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A copy machine makes

IrinaK [193]

This problem can be represented in a formula with fractions. To make it more simple, we are converting 5m30s to just pure seconds, which = 154 copies in 330 seconds.

You set up the formula like this: 154/330 = x/60. (it is x/60 because you want to know the x amount of copies per 60 seconds)

Cross multiply, you get 330x = 9240. Divide both sides by 330, you get:

x = 28 copies.

The final answer is 28 copies per minute

7 0

3 years ago

I AM GIVING BRAINLIEST

spayn [35]

Answer:

The first choice is correct

Step-by-step explanation:

$32.00 x 15% = 4.80. 32.00-4.80=27.20. Another way to work it is the first choice, subtracting 15% from 100% and then multiplying 85% by $32.00. Same answer

7 0

3 years ago

Whats is the measure of ikj

zimovet [89]

The is Answer: 202 because I answered it

6 0

3 years ago

Read 2 more answers

Solve and check thanks

navik [9.2K]

The solution is x=-1

6 0

3 years ago

Plz help me!!!

Nadya [2.5K]

To solve this we are going to use the exponential function: $f(t)=a(1(+/-)b)^t$
where
$f(t)$ is the final amount after $t$ years
$a$ is the initial amount
$b$ is the decay or grow rate rate in decimal form
$t$ is the time in years

Expression A
$f(t)=624(0.95)^{4t}$
Since the base (0.95) is less than one, we have a decay rate here.
Now to find the rate $b$ , we are going to use the formula: $b=|1-base|$ *100%
$b=|1-0.95|$ *100%
$b=0.05$ *100%
$b=$ 5%
We can conclude that expression A decays at a rate of 5% every three months.

Now, to find the initial value of the function, we are going to evaluate the function at $t=0$
$f(t)=624(0.95)^{4t}$
$f(0)=624(0.95)^{0t}$
$f(0)=624(0.95)^{0}$
$f(0)=624(1)$
$f(0)=624$
We can conclude that the initial value of expression A is 624.

Expression B
$f(t)=725(1.12)^{3t}$
Since the base (1.12) is greater than 1, we have a growth rate here.
To find the rate, we are going to use the same equation as before:
$b=|1-base|$ *100%
$b=|1-1.12|$ *100
$b=|-0.12|$ *100%
$b=0.12$ *100%
$b=$ 12%
We can conclude that expression B grows at a rate of 12% every 4 months.

Just like before, to find the initial value of the expression, we are going to evaluate it at $t=0$
$f(t)=725(1.12)^{3t}$
$f(0)=725(1.12)^{0t}$
$f(0)=725(1.12)^{0}$
$f(0)=725(1)$
$f(0)=725$
The initial value of expression B is 725.

We can conclude that you should select the statements:
- Expression A decays at a rate of 5% every three months, while expression B grows at a rate of 12% every fourth months.

- Expression A has an initial value of 624, while expression B has an initial value of 725.

8 0

3 years ago