Solution:
As given Square L M NO is dilated by a scale factor of two about the center of the square to create square L'M'N'O'.
Original line of Dilation = Along P Q
New Dilated line = P'Q'
As scale factor > 1
1. Image Size > Pre image size
2. The two images will be similar.
3. Length of Dilated Line P' Q' = 2 × Length of PQ
As you can see from the diagram drawn below, Dilated line P'Q' will contain the point P and Q.
All four points P,Q,Q',P' are collinear , lie in the same line.
Option (2) dilated line P'Q' will contain the points P and Q is true.
Answer:
If y(x-y)^2=x, then int1/(x-3y)dx is equal to (A) 1/3log{(x-y)^2+1} (B) 1/4log{(x-y)^2-1} (C) 1/2log{(x-y)^2-1} (D) 1/6 log{(x^2-y^2-1}
Step-by-step explanation:
Answer:
d=10u
Q(5/3,5/3,-19/3)
Step-by-step explanation:
The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane , then r will have the next parametric equations:
To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T
Substitute the value of in the parametric equations:
Those values are the coordinates of Q
Q(5/3,5/3,-19/3)
The distance from Po to the plane
Answer:
The equation for a is
The altitute is 101,428.57 feet
Step-by-step explanation:
You know that the relationship between ground temperature and atmospheric temperature can be described by the formula
t = -0.0035a +g
where:
- t is the atmospheric temperature in degrees Fahrenheit
- a is the altitude, in feet, at which the atmospheric temperature is measured
- g is the ground temperature in degrees Fahrenheit.
Solving the equation for a:
-0.0035a +g=t
-0.0035a= t - g
<u><em>The equation for a is </em></u><u><em></em></u>
If the atmospheric temperature is -305 °F and the ground temperature is 50 °F, then t= -305 °F and g= 50 °F
Replacing in the equation for a you get:
a= 101,428.57
<u><em>The altitute is 101,428.57 feet</em></u>