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stiks02 [169]
3 years ago
11

Plz help me thank you. 4 cm of the ice tray

Mathematics
2 answers:
disa [49]3 years ago
6 0
I had to do this and it’s fake, my whole class got it wrong even though we all put the right answer so ya
Readme [11.4K]3 years ago
4 0
I don’t know bud I need point so yea
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Follow me on brainy,and what is a+b+z+fg-y+up+f+r=​
pickupchik [31]

Answer:

Sure why not.

Step-by-step explanation:

7 0
3 years ago
Which expression is equivalent to (73)-2
padilas [110]

Answer:

73*-2 =-146 you have to multiply the 73 to the -2

Step-by-step explanation:

6 0
3 years ago
Solve for x on each of these questions please!
IgorLugansk [536]

Answer:

1. x = 10

2. x = 4

Step-by-step explanation:

I use the angle ABC method:

AB² + AC² = BC²

6² + 8² = x²

x = 10

AB² + AC² = BC²

3² + x² = 5²

x = 4

<em>H</em><em>O</em><em>P</em><em>E</em><em> </em><em>T</em><em>H</em><em>I</em><em>S</em><em> </em><em>H</em><em>E</em><em>L</em><em>P</em><em>S</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>H</em><em>A</em><em>V</em><em>E</em><em> </em><em>A</em><em> </em><em>N</em><em>I</em><em>C</em><em>E</em><em> </em><em>D</em><em>A</em><em>Y</em><em> </em><em><</em><em>3</em>

5 0
3 years ago
A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow
Mrrafil [7]

Answer:

\displaystyle \frac{54}{5405}.

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

\displaystyle \left(50\atop 5\right) = 2,118,760.

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

\displaystyle \left( 28\atop 2\right) = 378.

Number of ways to choose 3 green candies out of a batch of 8:

\displaystyle \left(8\atop 3\right)=56.

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168.

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}.

3 0
3 years ago
Can someone help me with exercise 121?
yanalaym [24]
It would be
(X^4+2X^2) -119(X^2-2X) then you would simplify
X^4+2X^2-119X^2-2X
So it would be
X^4+117X^2-2X=0
So basically X is undefined
8 0
3 years ago
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