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vovikov84 [41]
3 years ago
11

A quantity with an initial value of 850 grows continuously at a rate of 90% per

Mathematics
1 answer:
vredina [299]3 years ago
7 0

Answer:

5400

Step-by-step explanation:

so 90% of 850 is 765 (0.90x850)

765 is 10 years

so 1 year is 765/10 = 76.5 per year. So 76.5x59 = 4,513.5 add the initial ammount and you got 5,363.5 in the nearest hundredth you get 5400

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At an animal shelter, there are 15 dogs and 12 cats.
Bond [772]

Answer:

Dogs: 44.44 Cats: 55.556

Step-by-step explanation:

*this is <em>only </em>if you are allowed to use a calculator*

1. You have to add the total of the cats and dogs

Why? Here, let me put it this way. You have a test with two sections: math and science. If there are 2 Maths Questions and 2 Science questions, what percent of the test are math questions? You would not say 2/2 and say 100. You would say 2/4 because it is obvious. You do the same in this case but it is not as noticeable because you are dealing with bigger numbers.

15 + 12 = <u>27</u>.

2. Solve

<u>a. </u>There are 15 dogs and 12 cats that makes up 27 pets total.

So you have to do 12/27 for cats,

15/27 for dogs. The fractions equal one and It shows you how much of one is there compared to the total.

<u>b. </u><u><em>Simplify</em></u>

You can simplify these fractions:

for 12/27, divide the numerator and the denominator by 3.

You get 4/9

Divide 15/27 for dogs by 3 as well to get 5/9.

5+4 is 9, which means that I am doing the math correctly.

According to the calculator, 5/9 is 55.556, and 4/9 is 44.444

They should be accurate because there are more dogs than cats and the percentage for dogs are higher. Hence, this is correct.

Cats = 4/9 = 44.444%

Dogs = 5/9 = 55.556%

Hope this helps :)

-jp524

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2 years ago
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Answer:

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Step-by-step explanation:

5 0
2 years ago
I'm not sure where to start and how to differentiate this.
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The picture is not clear. let me assume y = (x^4)ln(x^3)

product rule :
d f(x)g(x) = f(x) dg(x) + g(x) df(x)

dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)
= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)

look at d[ln(x^3)/dx]
d[ln(x^3)/dx]
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end up with
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3 years ago
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