Artists such as Al Held, Bridget Riley etc are known for their abstract artwork or geometric artwork. Riley is famous for her optical artwork.
This art form is present among many cultures throughout the history both as decorative motifs and as art pieces themselves. Like in Islamic art, this art depicts religious figures. The optical art gives the viewer the impression of movement, hidden images, flashing and vibrating patterns. Basically it gives a disorienting effect.
Besides optical art, other abstract art forms are also famous. These imbibe a strong imagination and a sense of creativity in its viewers. This art form puts different colors, shapes, and textures together to create a finished piece that represents something in particular.
Yes, I appreciate this kind of art form because these forms are genuine and helps us understand the art patterns. Unlike the traditional form, these forms explore the relationships of forms/shapes and colors. Moreover, art always reflects culture. Therefore, abstract art is important because it is reflecting a culture that has been moving since last 2000 years.
Answer: I need the table to give the answer
Step-by-step explanation: copy and paste the table
Simplifying:
2n-21=0.25n
We move all terms containing n to the left and all other terms to the right.
+2n-0.25n=+21
We simplify left and right side of the equation.
+1.75n=+21
We divide both sides of the equation by 1.75 to get n.
n=12
![\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}} \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x} \\\\\\ (-2)^3=x\implies \boxed{-8=x}\\\\ -------------------------------\\\\ 0=\sqrt[3]{x}\implies \boxed{0=x}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dx%2B3x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B3%5Cleft%28%5Cfrac%7B2%7D%7B3%7Dx%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%20%20%5Cright%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B%5Ccfrac%7B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Csqrt%5B3%5D%7Bx%7D%2B2%5Cimplies%20-2%3D%5Csqrt%5B3%5D%7Bx%7D%0A%5C%5C%5C%5C%5C%5C%0A%28-2%29%5E3%3Dx%5Cimplies%20%5Cboxed%7B-8%3Dx%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A0%3D%5Csqrt%5B3%5D%7Bx%7D%5Cimplies%20%5Cboxed%7B0%3Dx%7D)
now, f(0) = 0, and f(-8) is an imaginary value or no real value.
now, f(-10) will also give us an imaginary value
and f(1) = 4
so, doing a first-derivative test on 0, is imaginary to the left and positive on the right, and before and after 1, is positive as well, so f(x) is going up on those intervals.
however, f(0) is 0 and f(1) is higher up, so the absolute maximum will have to be f(1), and we can use f(0) as a minimum, and since it's the only one, the absolute minimum.
the other two, the endpoint of -10 and the critical point of -8, do not yield any values for f(x).
15a^3+6a^2+25a+10
hope this helps!!!
