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oee [108]
2 years ago
5

26x32+2223-100 divided by 34

Mathematics
2 answers:
Roman55 [17]2 years ago
7 0

Answer:

54,395.17647058824

Step-by-step explanation:

Lady_Fox [76]2 years ago
4 0

Answer:

86 coma 91176470588235

explicación

espero que te sirva

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Please help me , fast
Vesnalui [34]

Answer:

0.3 with the hat ? the answer is first no, then it's a repeated decimal.

Step-by-step explanation:

5 0
2 years ago
Write an expression for "8 times z."
garri49 [273]

Answer: 8z 8(z)

Explanation: well 8 times z is simply 8(z)I don’t really think there is another way

5 0
2 years ago
Question 1 x - 2.5 = -3​
MAVERICK [17]

Answer:

x = -.5

Step-by-step explanation:

x - 2.5 = -3​

Add 2.5 to each side

x - 2.5+2.5 = -3+2.5​

x = -.5

8 0
2 years ago
How to do 30(c) ??????
adelina 88 [10]

Answer:

A kite with a 100 foot-long string is caught in a tree. When the full length of the string is stretched in a straight line to the ground, it touches the ground a distance of 30 feet from the bottom of the tree. Find the measure of the angle between the kite string and the ground.

17°

27°

63°

73°

Step-by-step explanation:

3 0
3 years ago
How many different ways are there to choose a subset of the set {1,2,3,4,5,6} so that the product of the members of the subset i
zvonat [6]

You have to pick at least one even factor from the set to make an even product.

There are 3 even numbers to choose from, and we can pick up to 3 additional odd numbers.

For example, if we pick out 1 even number and 2 odd numbers, this can be done in

\dbinom 31 \dbinom 32 = 3\cdot3 = 9

ways. If we pick out 3 even numbers and 0 odd numbers, this can be done in

\dbinom 33 \dbinom 30 = 1\cdot1 = 1

way.

The total count is then the sum of all possible selections with at least 1 even number and between 0 and 3 odd numbers.

\displaystyle \sum_{e=1}^3 \binom 3e \sum_{o=0}^3 \binom 3o = 2^3 \sum_{e=1}^3 \binom3e = 8 \left(\sum_{e=0}^3 \binom3e - \binom30\right) = 8(2^3 - 1) = \boxed{56}

where we use the binomial identity

\displaystyle \sum_{k=0}^n \binom nk = \sum_{k=0}^n \binom nk 1^{n-k} 1^k = (1+1)^n = 2^n

3 0
1 year ago
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