Acylinder and a cone have the same base and height. the cylinder can hold about 4,712 mc015-1.jpg of sand. jared says that the cone can hold about 1,178 mc015-2.jpg of sand. which explains whether jared is correct? jared is correct because the volume of the cone is less than the volume of the cylinder. the cone holds mc015-3.jpg less sand than the cylinder. jared is correct because the cone and the cylinder have the same base and height so the cone holds mc015-4.jpgof sand. jared is not correct because the cone and the cylinder have the same base and height so the cone holds mc015-5.jpg of sand. jared is not correct because the volume of the cone cannot be found without knowing the radius of the base and the height of the cone.

C

Step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1