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Elina [12.6K]
3 years ago
15

A corporation reported profits of approximately ​$600 million with approximately ​$11 million in revenues. Compare the profit to

revenue by writing as a fraction in lowest terms.
Mathematics
1 answer:
Alja [10]3 years ago
6 0

Answer:

For every $55 revenue, $3 is profit

Step-by-step explanation:

A corporation reports profit of 600,000,000

Revenue of 11,000,000,000

Therefore comparing the revenue to profit can be calculated as follows

= 600,000,000/11,000,000,000

= 6/110

= 3/55

Hence for every $55 revenue $3 out of it is the profit

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Given the function, f(x)= square root 3x+3+3, choose the correct transformation(s).
ki77a [65]

Answer:

3+3+3x

i think idk just getting points goog luck

Step-by-step explanation:

3 0
3 years ago
The parking rates at the hospital’s parking garage are 50 cents for the first hour and 25 cents for each additional hour. If Jua
Bad White [126]
The total charge for Juan in the parking spot would be $2.25.

Since we know one of those 8 hours is 50 cents, we can multiply .25 by 7 which would give us 1.75. Then we know that the first hour, as stated above, is going to be .50, so we can add that to our total which is 2.25.

Hope this helps!
3 0
3 years ago
Mathswatch: Factorise x^2-42^2
marta [7]

Answer:

x^{2} -1764

Step-by-step explanation:

1) Evaluate the exponent:

x^{2} -1*42^2\\x^{2} -1*1764

2) Multiply the numbers:

x^{2} -1*1764\\x^{2} -1764

6 0
3 years ago
Start with the basic function f(x)=2x. If you have an initial value of 1, then you end up with the following iterations. f(1)=2*
denpristay [2]
<span>f(x)=2x

f(1)=2*1=2

f^2(1)=2*2*1=4

f^3(1) =2*2*2*1=8

1. If you continue this pattern, what do you expect would happen to the numbers as the number of iterations grows?

I expect the numbers continue growing multiplying each time by 2.

Check your result by conducting at least 10 iterations.

f^4(1) = f^3(1) * f(1) = 8*2 = 16

f^(5)(1) = f^4(1) * f(1) = 16 * 2 = 32

f^6 (1) = f^5 (1) * f(1) = 32 * 2 = 64

f^7 (1) = f^6 (1) * f(1) = 64 * 2 = 128

f^8 (1) = f^7 (1) * f(1) = 128 * 2 = 256

f^9 (1) = f^8 (1) * f(1) = 256 * 2 = 512

f^10 (1) = f^9 (1) * f(1) = 512 * 2 = 1024

2. Repeat the process with an initial value of −1. What happens as the number of iterations grows?

f(-1) = 2(-1) = - 2

f^2 (-1) = f(-1) * f(-1) = - 2 * - 2 = 4

f^3 (-1) = f^2 (-1) * f(-1) = 4 * (-2) = - 8

f^4 (-1) = f^3 (-1) * f(-1) = - 8 * (-2) = 16

f^5 (-1) = f^4 (-1) * f(-1) = 16 * (-2) = - 32

As you see the magnitude of the number increases, being multiplied by 2 each time, and the sign is aleternated, negative positive negative positive ...
</span>
4 0
3 years ago
<img src="https://tex.z-dn.net/?f=y%20%3D%202x%20%2B%201%5C%5Cy%20%3D%202x%20%2B%205" id="TexFormula1" title="y = 2x + 1\\y = 2x
Gala2k [10]
What is the question?
8 0
3 years ago
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