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Paraphin [41]
3 years ago
15

Find the product 4/5x 7

Mathematics
2 answers:
GalinKa [24]3 years ago
7 0

Answer:

5 3/5  or 5.6

Step-by-step explanation:


prohojiy [21]3 years ago
3 0

Answer:

28/5

Step-by-step explanation:


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What is u equal to u=?
Jet001 [13]

Answer:

u= -449

Step-by-step explanation:

714=6(u+629)-366

<u>+366              +366</u>

1080=6u+3774

<u>-3774      -3774</u>

6u=-2694

<u>÷6   ÷6</u>

u=-449

6 0
4 years ago
What is 5+x^2=2x^2+13?
pshichka [43]
Take the root of both sides, and solve.

x = 2i sqrt 2, -2i sqrt 2.
4 0
3 years ago
So uh- people didnt do the question properly- i just need help with QUESTION 4
notka56 [123]

Answer:

A and C luv <3

Step-by-step explanation:

3 0
3 years ago
Olivia is buying bundles of wood for a campfire.The bundles of wood can't be split up.if each bundle costs $3,how many bundles c
victus00 [196]

Answer:

3    

1 $ left over

Step-by-step explanation:

       

                                 10$

         wood              wood                wood  

           3$                      3$                    3$

          she wood only buy 3 and have 1$ left

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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