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3 years ago

What is 19202 times 38392 divided by 2? get it right for brainliest!!!!!!!

Mathematics

2 answers:

g100num [7]3 years ago

8 0

368,601,592 is the correct answer

Salsk061 [2.6K]3 years ago

6 0

You multiply the two larger numbers first 19202 x 38392
and then divide that total by two: 368,601,592

You might be interested in

Through: (4, 3), parallel to x = 0

Gekata [30.6K]

Let P (x,y) be any point on the x-axis. Then clearly, for all position of P we shall the same ordinate 0 or, y = 0. Therefore, the equation of x-axis is y = 0. If a straight line is parallel and below to x-axis at a distance b, then its equation is y = -b.

8 0

3 years ago

There is a list of seven numbers. The average of the first four numbers is 5, and the average of the last four numbers is 8. If

sammy [17]

Answer:

Step-by-step explanation:

Sum of first 4 numbers= 4*5= 20

This includes 3 + 4th number

Sum of last 4 numbers= 4*8= 32

This includes 4th number + 3

Sum of 7 numbers= 7*(6+4/7)= 46

This includes 3+4th number +3

Number common to both sets= (20+32)- 46 = 6

3 0

3 years ago

Which expression is equivalent to 6(2x + 3) - 4(5x + 2)

AURORKA [14]

Answer:

A simplified/equivalent expression your equation would be - 8x + 10

Step-by-step explanation:

6(2x + 3) - 4(5x + 2)

Distribute:

= (6)(2x) + (6)(3) + (- 4)(5x) + (- 4)(2)

= 12x + 18 + - 20x + - 8

Combine Like Terms:

= 12x + 18 + - 20x + - 8

= (12x + - 20x) + (18 + - 8)

= - 8x + 10

Hope this helps!

8 0

3 years ago

What can you do to 40 to get 6? Thanks for your help whoever is answering I you for helping me!

Hoochie [10]

You would take away to get six

8 0

4 years ago

Read 2 more answers

In how many ways can 2 red, 2 black, 3 white and 2 blue balls be selected from 4 red, 3 black, 4 white and 8 blue balls? In how

Montano1993 [528]

From the total pool of colored balls, one can choose 2 reds, 2 blacks, 3 whites, and 2 blues in

$\dbinom42\cdot\dbinom32\cdot\dbinom43\cdot\dbinom82=6\cdot3\cdot4\cdot28=2016$

ways.

I'm assuming no ball of the same color is distinguishable from any other ball of the same color. So when I'm considering the possible arrangements, if I had lined up the ball as

red1 - black - red2 - ...

then this would be no different that

red2 - black - red1 - ...

So I now have 9 balls to arrange, which means there are $9!=362,880$ total possible permutations of them. But order among distinct colors is assumed to not matter. This means I have to divide the total number of permutations by the number of ways I could permute balls of the same color. Then there would be a total of

$\dfrac{9!}{2!\cdot2!\cdot3!\cdot2!}=7,560$

ways of arranging the balls I had selected.

5 0

3 years ago