Given end points of diameter,
(x1,y1)=(-3,-4)
(x2,y2)=(0,0)
Now,
the equation of circle is,
(x-x1)(x-x2)+(y-y1)(y-y2)=0
or, (x+3)(x-0)+(y+4)(y-0)=0
or, x^2 +3x +y^2 +4y =0
or, x^2 +y^2 +3x + 4y=0
which is in the form of x^2 +y^2 +2gx +2fy + c=0
where,
g=3/2
h=2
c=0
Now,

Answer:
1.) not safe
2.)

Step-by-step explanation:
1.) given the length of the ladder = 15ft.
and the height to the top of ladder when leaned against a wall is 14.8ft. This all forms a right triangle.
with what we are given we can solve for the angle it creates from the ground to the leaned ladder by using the SOHCAGTOA. Whike we do this keep in mind its not safe for a ladder to create an angle more than 70 degrees.
in this case if we are solving for the angle where the height is opposite we will use SOH. because we know the oposite and the hyp. Sin(theta) = opp/hyp


therefore not safe.
2.)
your givin 90 and 60. remember all interior angles add up to 180.
therefore 30 would be the unknown angle.
knowing that we use the <u>chart</u> at the top.
across from 30 is 6. so we put that by x. (remember we are doing this to find the height for our area of a triangle formula = base time height devide by 2.)
we need to find the height so since we kmow what x is we know what is across from 60 which is

so we plug that into our formula for area of triangle and u should get 18

Answer:
6.28km
Step-by-step explanation:
Hey,
We know that the circumfrence is the perimetert of a full circle. If we want to find the perimeter of a semicircle all we have to do is find the circumfrence and then divide it by 2 since we want to know the perimeter of half the circle.
Circumfrence is equal to 2πr. Where r is the radius and since we know the radius we an solve for the circumfrence.
2(π)(1km)=2kmπ.
The problem tells us to use 3.14 for pi so let's multiply 2 by 3.14.
6.28km is the answer!
Check the picture below.
so the quadrilateral is really just a parallelogram below a triangle, so let's simply get the area of each and sum them up.

Answer:
see below
Step-by-step explanation:
The line is AB with ↔ over it
The Plane requires 3 letters not all in a line
The plane name is ABD