what is the result of dividing 48a^3 + 32a^2 + 16a by 4a? A. 12a^2 + 8a + 4 B. 12a^2 + 4a + 8 C. 12a + 8 D. 12a^2 + 4a + 4 Simplify (2x-2y)(2y+8)=? A. 4xy-4y^2 B. 4xy-16x-4y^2-16y C. 4xy+16x-4y^2-16y D. 4xy+16x+4y^2+16y Simplify (-8q^3r^4s^2)^2=? A. 64q^9r^16s^4 B. -64q^6r^8s^4 C. 64q^6r^8s^4 D. -64q^9r^16s^4 What is the result if you divide -12x^8y^8 by 3x^4y^2? A. -4x^4y^6 B. 4x^2y^4 C. 4x^4y^6 D. -4x^2y^4

1. A. 12a^2 + 8a + 4

2. C. 4xy+16x-4y^2-16y

3. C. 64q^6r^8s^4

4. A. -4x^4y^6

Step-by-step explanation:

48a^3 + 32a^2 + 16a / 4a = (48a^3) / 4a + (32a^2) / 4a + (16a) / 4a = 12a^2 + 8a + 4.

So, the answer is A. 12a^2 + 8a + 4.

(2x-2y)(2y+8) = 4xy - 4y^2 + 16x -  16y.

So, the answer is C. 4xy+16x-4y^2-16y.

(-8q^3 * r^4 * s^2)^2 = (-8)^2 * q^6 * r^8 * s^4 = 64q^6r^8s^4.

So, the answer is C. 64q^6r^8s^4.

-12x^8y^8 / 3x^4y^2 = (-12 / 3) * (x^(8 - 4)) * (y^(8 - 2)) = -4x^4y^6.

So, the answer is A. -4x^4y^6.

Hope this helps!

17 1/3

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.