Answer:
The correct answer is :
1. Line PQ (One line PQ).
Step-by-step explanation:
The first step to solve this question is to draw the plane A with the points P and Q lying on it.
We know that given two different points there is only one line that contains this two different points.
Let's analyze each option.
''2. Lines PQ and QP''
This option is wrong because there aren't two different lines. In fact it is only one line that can be named line PQ or line QP.
''3. The 2 lines PQ and QP plus another line that does not lie in plane A.''
This option is assuming that exist three lines that contain P and Q. This option is also wrong.
''1. Line PQ''
This option is correct. It will be clarify with the drawing I will attach.
''We can't name them all!''
This option is assuming that exist infinite lines that contain P and Q. This option is wrong.
In the drawing I call the line that contains P and Q as line L.
Given that P and Q lie in plane A necessarily the line L must lie on the plane A.
I think it’s either C or D
Check the picture below
it can't be -3... since is width unit, so it has to be the other
and recall that length = 2w - 5
Answer:
4 ft downwards
Step-by-step explanation:
old height off the ground :
7^2+b^2=25^2
49+b^2=625
b^2=576
b=24
new height off the ground :
7+8=15
15^2+ b^2 =25^2
225+b^2 = 625
b^2 = 400
b=20
difference between heights :
24 - 20 = 4
Answer:
1. A = 59
2. A = 43
Step-by-step explanation:
If we have a right triangle we can use sin, cos and tan.
sin = opp/ hypotenuse
cos= adjacent/ hypotenuse
tan = opposite/ adjacent
For the first problem, we know the opposite and adjacent sides to angle A
tan A = opposite/ adjacent
tan A = 8.8 / 5.2
Take the inverse of each side
tan ^-1 tan A = tan ^-1 (8.8/5.2)
A = 59.42077313
To the nearest degree
A = 59 degrees
For the second problem, we know the adjacent side and the hypotenuse to angle A
cos A = adjacent/hypotenuse
cos A = 15.3/21
Take the inverse of each side
cos ^-1 cos A = cos ^-1 (15.3/21)
A = 43.23323481
To the nearest degree
A = 43 degrees
