Answer:
x=1
Step-by-step explanation:
According to your typed question,
f(x)=2x-3
Let f(x)=y
y=2x-3
Now,
Interchanging the positions of x and y
x=2y-3
x+3=2y
x+3/2=y
f'(x)=x+3/2
Then,
f(x)=f'(x)
2x-3=x+3/2
2(2x-3)=x+3
4x-3=x+3
4x-x=3+3
3x=3
x=3/3
x=1
According to your image question,
f(x)=x/2x-3
f(x)=f'(x)
Now,
Let y=f(x)
y=x/2x-3
y(2x-3)=x
2xy-3=x
2xy-x=3
x(2y-1)=3
2y-1=3x
2y=3x+1
y=3x+1/2
f'(x)=3x+1/2
Then,
f(x)=f'(x)
x/2x-3=3x+1/2
2x=6xsq+2x-9x+3
2x=6xsq+7x+3
solve for x ok
Answer:
Attached diagram A'B'C'D'
Step-by-step explanation:
Given is a quadrilateral ABCD. It says to draw a dilated version with a scale factor 2/3.
We see that scale factor is less than 1 which means it shrinks the image to a smaller one.
To draw a scaled copy, we need to find the lengths of its sides.
To do so, we can draw the diagonals AC & BD, and they intersect at origin O(0,0) such that OA= -2, OB= 2, OC= 4, OD= -4.
Applying a scale factor of 2/3, we get OA' = -4/3, OB' = 4/3, OC' = 8/3, OD' = -8/3.
So we have attached a scaled copy A'B'C'D' of quadrilateral ABCD with a scale factor 2/3.
Answer:
whats the question
Step-by-step explanation:
will be edited when answered
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
The answer is, there are 320 possible combinations of outfits
