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scZoUnD [109]
3 years ago
9

Consider this expression and the steps to evaluate it.

Mathematics
2 answers:
Luda [366]3 years ago
6 0

Answer:

A=6

B=3

C=64

D=64

Hope This helps

vladimir2022 [97]3 years ago
5 0

Answer:

A=6

B=3

C=64

D=64

Step-by-step explanation:

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The sum of 4 consecutive integers is 182. W hat is the sum of the middle two numbers?
Kamila [148]

Answer:

91

Step-by-step explanation:

Let

x = first number

x+1 = second number

x+2 = third number

x+3 = fourth number

Total = 182

x + x + 1 + x + 2 + x + 3 = 182

4x + 6 = 182

Subtract 6 from both sides

4x + 6 - 6 = 182 - 6

4x = 176

Divide both sides by 4

x = 176 / 4

= 44

x = 44

x = 44

x+1 = 44 + 1 = 45

x+2 = 44 + 2 = 46

x+3 = 44 + 3 = 47

Sum of the two middle numbers = 45 + 46

= 91

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3 years ago
The lunch lady has 6 lb of lasagna left over if she makes one and a half pound serving size how many servings of lasagna can she
AfilCa [17]
She can make 4 servings with 0 left over. 6/1.5=4
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3 years ago
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Jasmine makes $36,000 per year as a teaching assistant. Her monthly check amount is: *
joja [24]
It is b jjajsjsjshsjsjsj
3 0
3 years ago
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What is the ANSWER to this problem
Vadim26 [7]

Answer:


Step-by-step explanation:


8 0
4 years ago
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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
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