6(x-1)=30 There is my

Allisa [31]

Not sure what you are asking for, but,

Here is an example of a JRU (Join Result Unknown) word problem:

There were _____ kids on the playground. ____ more kids came onto the playground. How many kids are on the playground?

Here is an example of a JCU (Join Change Unknown) word problem:

There were ____ kids on the playground. Some more kids came on the playground. Now there are ____ kids on the playground. How many kids came on the playground?

Here is an example of a JSU (Join Start Unknown) word problem:

Some kids were on the playground. ____ kids came on the playground. Now there are ____ kids on the playground. How many kids were on the playground at the beginning?

8 0

3 years ago

A line in a graph that runs horizontally (left-right) through zero is

Rom4ik [11]

The x-axis brainliest plz :)

3 0

3 years ago

For accounting purposes, the value of assets (land, buildings, equipment) in a business are depreciated at a set rate per year.

Scrat [10]

Answer:

So, option d is correct i.e,

V₀ = $393,000, b = 0.85, and the value after 7 years is $125,986.80

Step-by-step explanation:

The formula given is V(t)=V₀(b)^t

The value of V₀ (the actual worth) is:

V₀ = $393,000

The value of b is :

b= (1-15%) = (1-0.15) = 0.85

Value of assets after 7 years is:

t= 7, V₀ = $393,000, b=0.85

putting values in formula:

V(t) = Vo(b)^t.

V(7)= $393,000 * (0.85) ^ 7

V(7)= $393,000 * (0.320)

V(7)= $125986.80

So, option d is correct i.e,

Vo = $393,000, b = 0.85, and the value after 7 years is $125,986.80

4 0

3 years ago

6x + ____ = 10 Please find the blank.

andre [41]

Answer:

6x +4=10

I think this is answer

7 0

2 years ago

In 1990, the rate of change of the world population was approximately 0.09125 billion per year (or approximately 1 million peopl

babunello [35]

Answer: Option A

$P = 5.3 + 0.09125t$

Step-by-step explanation:

Note that for the initial year, 1990, the population was 5.3 billion.

The exchange rate is 0.09125 billion per year. In other words, each year there are 0.09125 billion more people.

In year 2 there will be 0.09125 * 2 billion people

In year 3 there will be 0.09125 * 3 billion people

In year 4 there will be 0.09125 * 4 billion people

In year t there will be 0.09125 * t billion of people

So the equation that models the number of people that there will be as a function of time is:

$P = P_0 + rt$

Where $P_0$ is the initial population

$P_0 = 5.3$ billion

r is the rate of increase

$r = 0.09125$ billion per year

finally the equation is:

$P = 5.3 + 0.09125t$

The correct answer is option A.

6 0

4 years ago

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