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patriot [66]
3 years ago
9

Simplify (2^8*3^-5*6^0)^-2 * (3^-2\2^3)-^4 * 2^28=

Mathematics
1 answer:
marin [14]3 years ago
3 0
This is pretty hard but um in decimal form it would make 0.9010162353515625, but to simplify it the answer would be 0.9
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Please answer this in two minutes
Vladimir79 [104]
Not a triangle because the two smaller numbers do not add up to a bigger number than 15
3 0
3 years ago
Read 2 more answers
The population of a town is 13,000. It decreases at a rate of 5% per year. In about how many years will the population be less t
just olya [345]

This is an exponential decay problem.

Using the equation Y = a *(1-rate)^time

where Y is the future value given as 12,000 and a is the starting value given as 13,000.

The rate is also given as 5%.

The equation becomes:

12,000 = 13,000(1-0.05)^x

12,000 = 13,000(0.95)^x

Divide each side by 13000:

12000/13000 = 0.95^x/13000

12/13 = 0.95^x

Use the natural log function:

x = ln(12/13) / ln(0.95)

x = 1.56 years. ( this will equal 12,000

Round to 2 years it will be less than 12000.

3 0
3 years ago
Bill has a rectangular garden that is 3 1/2 feet by 4 1/2 feet. A bag of fertilizer covers 5 1/4 square feet. What is the minimu
sveta [45]

Answer:

3

Step-by-step explanation:

3.5*4.5=15.75 ft^2

15.75/5.25=3 bags

8 0
3 years ago
When the following expression is written in simplest form, what is the coefficient of the variable term? -2.5y + 8y - 10.5y
LUCKY_DIMON [66]

Answer:

Option 1 is correct.

-5

Step-by-step explanation:

Given:

The given expression is.

=-2.5y+8y-10.5y

Write the given expression in simplest form.

=-2.5y+8y-10.5y

Add negative term of y.

=-13y+8y

=-5y

The simplest form of given expression is =-5y

Therefore, the coefficient of the variable term is -5

7 0
3 years ago
Of the following fractions: 9/19, 5/11, 7/15, and 11/23, which is the largest?
dusya [7]

Answer: 11/23

Step-by-step explanation:

I used my calculator to convert those fractions into decimal form and found the biggest number. It was about 0.4789.

5 0
3 years ago
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