Answer:
f(3/2) = 13√13 ≈ 46.872167
Step-by-step explanation:
Put 3/2 where x is and evaluate.
f(3/2) = 13^(3/2) = 13^(1 +1/2) = 13√13
f(3/2) ≈ 46.872167
I hope this helps you
x'=1 y'=4 slope = -3
slope. (x'-x)=y'-y
-3. (1-x)=4-y
-3+3x=4-y
y=4+3-3x
y=7-3x
Answer:
Yuri is not correct.
Step-by-step explanation:
Given expression is q(x) = 6x³ + 19x² - 15x - 28
If 'a' is a root of the given function, then by substituting x = a in the expression, q(a) = 0
Similarly, for x = 
,
= 
= 
= 
= 
= 
≠ 0
Therefore, Yuri is not correct. x = can not be a root of the given expression.
Answer:
78
Step-by-step explanation:
The tricky part of this is figuring out how to assign the unknowns. We are told that we are working with two consecutive even integers. Consecutive means "next to" or "in order" and sum means to add. If we use 2 and 4 as examples of our 2 consecutive even integers and assign x to 2, then in order to get from 2 to 4 we have to add 2. So the lesser of the 2 integers is x, and the next one in order will be x + 2. (2 and 4 are just used as examples; they mean nothing to the solving of this particular problem. You could pick any 2 even consecutive integers and find the same rule applies. All we are doing here with the example numbers is finding a rule for our integers.) Now we have the 2 expressions for the integers, we will add them together and set the sum equal to 158:
x + (x + 2) = 158
The parenthesis are unnecessary since we are adding, so when we combine like terms we get
2x + 2 = 158 and
2x = 156 and
x = 78
That means that the lesser of the 2 integers in 78, and the next one in order would be 80, and 78 + 80 = 158
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:
A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?
You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked
2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.
b - 1,000,000/2500 = 400
400 packages are defective each year
