Answer: D
Step-by-step explanation:
Using the equation plot in the coordinates of the points to see if they fit the equation.
5x-3y < 30
A) 5(-3) -3(5) < 30
-15 -15 < 30
-30 < 30 This is true
B) 5(0) -3(0) < 30
0 - 0 < 30
0< 30 This is true
C) 5(-5) - 3(3) < 30
-25 - 9 < 30
-34 < 30 This is also true
D) 5(3) - 3(-5) < 30
15 + 15 < 30
30 < 30 This is not true because 30 is not greater that 30
I think it’s a but I don’t know if it’s right
The equation to this graph is Y=(x-2)^2-5
Showing that
H=-2
K=-5
Answer: At noon the temperature outside of Hector's home was 4 degree warmer than the temperature outside of gale's home.
Step-by-step explanation:
Here, x represents the number of hours after noon while g(x) and f(x) represents the temperature of gale's and hector's home respectively in x hours.
Since, at noon, x = 0
And, by the table g(x=0) = -1
That is, the temperature of gale's home is -1° F at noon.
And, by the given graph, f(x=0) = 3
That is, the the temperature of hector's home is 3° F at noon.
And, f(0)-g(0) = 3-(-1) = 4°F
That is, the difference between the temperature of hector's home at noon and the temperature of gale's home at noon is 4°F.
The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.
y² = x
Take the derivative of both sides with respect to x, assuming y = y(x) :
2y dy/dx = 1
dy/dx = 1/(2y)
Solve for y when dy/dx = 1 :
1 = 1/(2y)
2y = 1
y = 1/2
When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.
This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :
x = y² ⇒ y = y² + 3 ⇒ y² - y + 3 = 0
has no real solution for y.