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Komok [63]
2 years ago
12

Farad wrote the expression 32 - 8f to represent

Mathematics
2 answers:
motikmotik2 years ago
7 0
Look at this and you will be surprised on how it helps many like you that’s a key answer

barxatty [35]2 years ago
5 0
Im not sure but I will try to figure it out
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Mrs. Chu's famous peanut butter cookies call for 1 cup of peanut butter for every 1/2 of a cup of oil. Today, she wants to make
lesantik [10]

Answer:

she should use 2 cups of peanut butter

Step-by-step explanation:

to know the answer to that

use this equation (pb is peanut butter &o is oil)

1cup of pb=1/2 cup of o

?=1 cup of o

1×1÷1/2= 1×1×2/1=2co cups of pb

5 0
3 years ago
Hozumi recorded the daily high temperatures for an 18-day period in the table shown please I need this really badly
mr Goodwill [35]

Answer:

the answer is letter A

Step-by-step explanation:

7 0
3 years ago
Find the slope of the line.<br><br> The slope is ___
dimulka [17.4K]
The slope would be 2/2
3 0
3 years ago
Read 2 more answers
What is the gradient between (0,3) and (1,8).
trasher [3.6K]

Step-by-step explanation:

Given

(x1 , y1) = ( 0 , 3)

(x2 , y2) = ( 1 , 8)

Now

Gradient =

\frac{y2 - y1}{x2 - x1}

=  \frac{8 - 3}{1 - 0}  \\  =  \frac{5}{1}  \\  = 5

Hope it will help :)

5 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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