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3 years ago

Rewrite the following expression using only addition: 4x-2y-8

Mathematics

1 answer:

geniusboy [140]3 years ago

7 0

Step-by-step explanation:

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There is a sales tax of $25 on an item that costs $310 before tax. the sales tax on a second item is $10 . how much does the sec

Maksim231197 [3]

Answer:

$124

Step-by-step explanation:

If the tax value of a item that costs $310 before tax is $25, we can calculate the tax percentage by dividing the tax costs by the item value without taxes:

tax = 25/310 = 0.0806

So, for the second item, if the tax costs were $10, we can calculate the item value (this value will be called X) with the following equation:

tax = 10/X = 0.0806 -> X = 10/0.0806 = $124

Another way of solving this problem, is by rule of three:

tax of 25 for an item of $310,

tax of 10 for an item of $X

25/10 = 310/X -> X = 310*10/25 = $124

5 0

3 years ago

To stay within your budget,the area of the house and the garage combined is at most 3000 square feet.The area of the garage is 5

navik [9.2K]

Answer:

h ≤ 3000 ft² - 528 ft²

h ≤ 2472 ft²

Step-by-step explanation:

Area of house and garage is at most 3000 ft²

Area of garage = 528 ft²

Area of house (h) =?

Area of garage + Area of house ≤ 3000 ft²

528 ft² + h ≤ 3000 ft²

h ≤ 3000 ft² - 528 ft²

h ≤ 2472 ft²

6 0

3 years ago

What equation best models this data?(use y to represent the population of rabbits and t to represent the year, assuming that 201

liraira [26]

If we see the data closely, a pattern emerges. The pattern is that the ratio of the population of every consecutive year to the present year is 1.6

Let us check it using a couple of examples.

The rabbit population in the year 2010 is 50. The population increases to 80 the next year (2011). Now, $\frac{80}{50}=1.6$

Likewise, the rabbit population in the year 2011 is 80. The population increases to 128 the next year (2012). Again, $\frac{128}{80}$ $=1.6$

We can verify the same ratio with all the data provided.

Thus, we know that the population in any given year is 1.6 times the population of the previous year. This is a classic case of a compounding problem. We know that the formula for compounding is as:

$F=P\times r^n$

Where $F$ is the future value of the rabbit population in any given year

$P$ is the rabbit population in the year "0" (that is the starting year 2010) and that is 50 in this question. (please note that there is just one starting year).