Answer:
87 packages
Step-by-step explanation:
First we need to find the volume of the cone-shaped vase.
The volume of a cone is given by:
V_cone = (1/3) * pi * radius^2 * height
With a radius of 9 cm and a height of 28 cm, we have:
V_cone = (1/3) * pi * 9^2 * 28 = 2375.044 cm3
Each package of sand is a cube with side length of 3 cm, so its volume is:
V_cube = 3^3 = 27 cm3
Now, to know how many packages the artist can use without making the vase overflow, we just need to divide the volume of the cone by the volume of the cube:
V_cone / V_cube = 2375.044 / 27 = 87.9646 packages
So we can use 87 packages (if we use 88 cubes, the vase would overflow)
No thanks. Thanks for the point
This is in slope-intercept form, so you can think of it as being y=mx+b. Invert m (the slope) and flip its sign, and then substitute the x and y values in and solve for b. The answer is y=7/2x-20.
(y-x) (2+3)= (yx)(5)
But am not sure it could be =(2-3) (y+x) = (1) (yx)
Answer: D) 13y^25 and 2y^25
Like terms involve the same variables, and each of those variables must have the same exponents.
Another example of a pair of like terms would be 5x^3y^2 and 7x^3y^2. Both involve the variable portion "x^3y^2" which we can replace with another variable, say the variable z. That means 5x^3y^2 becomes 5z and 7x^3y^2 becomes 7z. After getting to 5z and 7z, it becomes more clear we have like terms.
