Hi there!


We can calculate dy/dx using implicit differentiation:
xy + y² = 6
Differentiate both sides. Remember to use the Product Rule for the "xy" term:
(1)y + x(dy/dx) + 2y(dy/dx) = 0
Move y to the opposite side:
x(dy/dx) + 2y(dy/dx) = -y
Factor out dy/dx:
dy/dx(x + 2y) = -y
Divide both sides by x + 2y:
dy/dx = -y/x + 2y
We need both x and y to find dy/dx, so plug in the given value of x into the original equation:
-1(y) + y² = 6
-y + y² = 6
y² - y - 6 = 0
(y - 3)(y + 2) = 0
Thus, y = -2 and 3.
We can calculate dy/dx at each point:
At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.
At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.
If you're using the app, try seeing this answer through your browser: brainly.com/question/3000586——————————
The answer is option
D) r < 5 or r > – 1.
I'm going to graph each inequality below on a number line.
A) r > 5 or r > – 1.

The result is found just by joining those two intervals together. Actually that compound inequality only implies
r > – 1which does not represent all real numbers.
—————
B) r > 5 or r < – 1.

Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.
—————
C) r < 5 or r < – 1.

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.
—————
D) r < 5 or r > – 1.
Now
all real numbers are included in the union. So this is the right choice.
Answer:
option D) r < 5 or r > – 1.
I hope this helps. =)
Answer:
19
Step-by-step explanation:
given:
f(x)=3x²-8
g(x)=4x+1
g(-1) = 4(-1)+1
g(-1) = -4 + 1
g(-1) = -3
hence
f[g(-1)]
= f(-3)
= 3(-3)²-8
= 3(9) - 8
= 19
Put a point at -9 because that is the y-intercept.then from the point -9 count up 5 and count over 3.
Answer:
Standard Deviation = 5.928
Step-by-step explanation:
a) Data:
Days Hours spent (Mean - Hour)²
1 5 61.356
2 7 34.024
3 11 3.360
4 14 1.362
5 18 26.698
6 22 84.034
6 days 77 hours, 210.834
mean
77/6 = 12.833 and 210.83/6 = 35.139
Therefore, the square root of 35.139 = 5.928
b) The standard deviation of 5.928 shows how the hours students spend outside of class on class work varies from the mean of the total hours they spend outside of class on class work.