Answer:
6x
Step-by-step explanation:
because in front of x^3 there is 6x and also the qn has asked to find the coefficient of x^3
Multiply every number from the small octagon by 7 then add the products to get 238.
For problems like this you can move the y's on the bottem up top by fliping the sign of the exponent (in this case 7 to -7) and MULTIPLYING it with EVERYTHING on top, then because of properties of multplication you can add the exponents to combine them into one term (in this case add 3 and -7 to get -4)
If any of this does not make sense let me know and ill try to clarify better
Each as a fraction 42/100 35/100 18/100 100/100
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
?