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3 years ago

True or false 16.-84.-184-284 is an arithmetic sequence? O True O False

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

6 0

Answer:

false

plz tell me if im wrong i hope i helped

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How many 1/2 unit cubes can fit in a a prism with a volume of 6 cubic units

AlekseyPX

Answer:

48 cubes

Step-by-step explanation:

step 1

Find the volume of one 1/2 unit cube

The volume of the cube is given by

$V=b^3$

where

b is the length side of the cube

we have

$b=\frac{1}{2}\ units$

substitute

$V=(\frac{1}{2})^3$

$V=\frac{1}{8}\ units^3$

step 2

Divide the volume of the prism by the volume of one cube

$6:\frac{1}{8}=6(8)=48\ cubes$

3 0

3 years ago

Math 8th grade test help help

Rufina [12.5K]

Answer:

Step-by-step explanation:

D has a repeating x value. Functions are only supposed to have one input for everyone output. ( no repeating x values)

5 0

3 years ago

Ray created the list of factors for 48 below.

brilliants [131]

Answer:

Step-by-step explanation:

also why is 78 there? a factor means in that range

3 0

3 years ago

Read 2 more answers

Someone please help even if u just answer one of the questions :)

Advocard [28]

1. E
2(a+2b+3c)= 2a+4b+6c you will distribute the 2 into a, b, and c which you will multiply.
2 x a = 2a
2 x 2b = 4b
2 x 3c = 6c

3 0

3 years ago

Rita is selecting a sandwich at the deli. The deli has four types of meat, three types of cheese and two types of bread. A delux

zalisa [80]

Answer:

24 possible dishes

Step-by-step explanation:

Given

Total

$Meat = 4$

$Cheese = 3$

$Bread = 2$

Selection

$Meat = 1$

$Cheese =2$

$Bread = 1$

Required

Determine the number of possible combo

Rita wants to select 1 of 4 meat types,

So, the number of selections is: $^4C_1$

Rita wants to select 2 of 3 cheese types,

So, the number of selections is: $^3C_2$

Rita wants to select 1 of 2 cheese types,

So, the number of selections is: $^2C_1$

The total number of selections is:

$Total = ^4C_1 * ^3C_2 * ^2C_1$

Using

$^nC_r = \frac{n!}{(n-r)!r!}$

We have:

$Total = \frac{4!}{(4-1)!1!} * \frac{3!}{(3-2)!2!} * \frac{2!}{(2-1)!1!}$

$Total = \frac{4!}{3!1!} * \frac{3!}{1!2!} * \frac{2!}{(1!1!}$

$Total = \frac{4*3!}{3!*1} * \frac{3*2!}{1*2!} * \frac{2*1}{1*1}$

$Total = 4* 3 * 2$

$Total = 24$

<em>Hence, there are 24 possible dishes</em>

4 0

3 years ago

Read 2 more answers