Answer:
Osvoldo did not meet his goal
Step-by-step explanation:
To know the percentage of carbohydrates that Osvoldo ate from grains with respect to the total, we simply have to divide the amount of carbohydrates he got with the grains by the total
55 / 220 = 0.25
then to express it as a percentage this number has to be multiplied by 100
0.25 * 100 = 25%
Osvoldo wanted to reach 30% and only got 25%
25% < 30%
The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1
If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1
If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)
If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)
however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1
If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1
and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)
Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.
The <em><u>correct answer</u></em> is:
At least 7 times.
Explanation:
Since 10% of the water in the jug is removed each time, this means that 100-10=90% of the water remains.
After two times, we would have 0.9(0.9) = 0.81 = 81% of the water; each time she pours, the number of times 0.9 is multiplied by itself changes. This makes this an exponential inequality (it is not an equation, as we want "less than half of the water"):
To solve this, we will use logarithms:
