Pablo Picasso’s Head of a Women, is an example of a(n) ______________.

Mathematics

2 answers:

lina2011 [118]3 years ago

8 0

Answer:

Sculpture in the round.

Step-by-step explanation:

Option A, sculpture in the round, is the right answer.

This is Picasso’s initial Cubist sculpture illustrations the master made of his girlfriend Fernande Olivier in the year 1909. After the return of the couple from a tour to Spain Picasso created the bust in Paris. As in his old Cubist pictures, the frame of her sculpted head is faceted into miniature pieces. Meant to be viewed in the round, the production switches form when seen from various corners. Therefore, it is clear that Pablo Picasso’s "Head of a Women" was a sculpture in the round.

romanna [79]3 years ago

8 0

<h3><u>Answer:</u></h3>

Pablo Picasso’s Head of a Women, is an example of a(n) Sculpture in the round.

<h3><u>Step-by-step explanation:</u></h3>

Pablo Picasso's Head of a woman:

This was a sculpture which was made by Pablo Picasso. It was Picasso's first cubist sculptures and one of more than sixty Cubist paintings,sculptures and drawings the artist made of his lover.

It was made in the year 1909.

He modeled the bust after the couple returned from summer trip to spain.

The composition changes when the sculpture is viewed from different angles.

Hence, option: a is the correct answer.

a. sculpture in the round.

You might be interested in

The area of the scale model of a playground is 6 square yards. The scale model is enlarged by a scale factor of 3 to create the

REY [17]

Answer:

B: 54 square feet

Step-by-step explanation:

There are 3 feet in 1 yard. The scale model is 6 square yards.

Multiply 6 by 3 to get the amount of square yards.

There are 18 square yards in the actual playground.

Since there are 3 feet in a yard, multiply 18 by 3.

This gives you your answer:

54 square feet

6 0

3 years ago

Read 2 more answers

Using a directrix of y = −2 and a focus of (1, 6), what quadratic function is created? Question 3 options:

deff fn [24]

If your directrix is a "y=" line, that means that the parabola opens either upwards or downwards (as opposed to the left or the right). Because it is in the character of a parabola to "hug" the focus, our parabola opens upwards. The vertex of a parabola sits exactly halfway between the directrix and the focus. Since our directrix is at y = -2 and the focus is at (1, 6) AND the parabola opens upward, the vertex is going to sit on the main transversal, which is also the "line" the focus sits on. The focus is on the line x = 1, so the vertex will also have that x coordinate. Halfway between the y points of the directrix and the focus, -2 and 6, respectively, is the y value of 2. So the vertex sits at (1, 2). The formula for this type of parabola is $(x-h)^2=4p(y-k)$ where h and k are the coordinates of the vertex and p is the DISTANCE that the focus is from the vertex. Our focus is 4 units from the vertex, so p = 4. Filling in our h, k, and p: $(x-1)^2=4(4)(y-2)$ . Simplifying a bit gives us $(x-1)^2=16(y-2)$ . We can begin to isolate the y by dividing both sides by 16 to get $\frac{1}{16}(x-1)^2=y-2$ . Then we can add 2 to both sides to get the final equation $f(x)=\frac{1}{16}(x-1)^2+2$ , choice 4 from above.