Answer:
Position of the submarine now relative to the water surface 
Step-by-step explanation:
It says that the position of a submarine relative to the water surface was -32 1/4 feet . A downward navigational maneuver increased it's depth by 15 1/2 feet.
Now we need to find about what is the position of the submarine now relative to the water surface.
Downward navigation means we 15 1/2 is negative
so we will just add both values to get the final answer:
Position of the submarine now relative to the water surface = 
Check the slopes
the first 2 points have slope (3-1) / (5-1) = 2/4 = 1/2
points 2 and 3 slope = 4/2 = 2
3 and 4:- slope = -2/-4 = 1/2
4 and 1:- slope = 4/2 = 2
so we have 2 pairs of opposite sides which are parallel
looks like a parallelogram or a rhombus
if adjacent sides are equal its a rhombus
length betwee pts 1 and 2 = sqrt(2^2 + 4^2) = sqrt20
.. ... ... .. .. .. ...2 and 3 = sqrt(4^2 + 2^2) = sqrt20
so it is a rhombus
This is the picture I am not very good at proofs but hopefully someone can solve it.
Answer:
w => -4
w =<9
Step-by-step explanation:
We have been given an image of a circle. We are asked to find the measure of angle JKL.
We can see that angle JML is a central angle of our given circle, so measure of angle JML will be equal to measure of arc JL. So measure of angle JML will be 60 degrees.
We know that tangent of a circle is perpendicular to radius.
We can see that tangent KJ is tangent to circle at point J and tangent KL is tangent to circle at point L. This means that measure of angle KJM and angle KLM will be 90 degrees each.
We can see that JKLM is a quadrilateral. We know that all angles of a quadrilateral add up-to 360 degrees, so we can set an equation as:
Therefore, measure of angle JKL is 120 degrees and option B is the correct choice.
