Answer:White Paintings
Step-by-step explanation:
In 1951, Robert Rauschenberg painted some stretched canvanses a plain, solid white, leaving minimal roller marks. Each of his works consist of different number of panel iterations ( one to seven panels) which are collectively known as 'the white paintings'.
The measure of arc EF = 146 degrees because its central angle = 146 degrees.
You know how there are 4 different areas in the graph, rotating it 180 degrees from that position will make it be in quadrant 2 (top left). in order to graph the ponts, since it is right across, then you just have to flip the shape around. like how you take a selfe, it flips the capture image around
You need to use Pythagoras’ Theorem.
A² + B² = c² and c is the hypotenuse of the right angled triangle.
The height of the tower (5 feet) is A and the distance from the end of the cable and the base of the tower (12 feet) is B. The length of ONE cable is c. So:
A² + B² = c²
5² + 12² = c²
25 + 144 = c²
169 = c²
13 = c. This is the length of one cable.
3 x 13 = 39 therefore the total length of the cables is 39 feet.
25
Between the probability of union and intersection, it's not clear what you're supposed to compute. (I would guess it's the probability of union.) But we do know that

For parts (a) and (b), you're given everything you need to determine

.
For part (c), if

and

are mutually exclusive, then

, so

. If the given probability is

, then you can find

. But if this given probability is for the intersection, finding

is impossible.
For part (d), if

and

are independent, then

.