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Gekata [30.6K]
3 years ago
6

Find the magnitude & directionof the vector JK given point j(1,0) & k(-2,3)

Mathematics
1 answer:
julia-pushkina [17]3 years ago
7 0
The distance between two point:
L=sqrt[(1-(-2))^2+(0-3)^2]=sqrt(18)=3sqrt(2) - <span>magnitude
</span>About <span>direction. Angle is 45</span>°, both point are on the line y=-x+1

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Please helppp! I don’t know the answer
disa [49]

Answer:

250x+140=y

Step-by-step explanation:

so amount earned for 10 hours of work is 250

8 0
3 years ago
The larger of two numbers is five more than twice the smaller number. The sum of the two numbers is 38. Find the numbers.
Damm [24]

Answer:

I would set this up as:

let: lesser number = x

     greater number = 2x + 5

Restating the word problem:

lesser number + greater number = 38

        x           +  2x + 5             = 38

Solving:  3x + 5 = 38

                  - 5     -5

             ---------------

              3x      =  33

              ---          ---

               3            3

               x        =  11

Substitution for the greater number:

        2(11) + 5

          22    + 5

              27

8 0
3 years ago
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What is the perimeter of a rectangle with sides of 4 inches by 8 inches?​
olasank [31]
80
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3 years ago
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The length of a rectangle is twice its width.
zimovet [89]

Answer:

idc

Step-by-step explanation:

3 0
3 years ago
the mean height of a group of 500 nonsmoking college students is 74 inches and the standard deviation is 5 inches. what is the p
Novosadov [1.4K]
Based on the given description above, I have analyzed it and come up with a solution to get the probability if in a random sample of 25 students from this said group and the average height is between 73 and 75 inches. So if you calculate it, it will be like this: 
<span>2P(Z<2)−1</span> To find the probability of Z, use the normal distribution table.
The value for Z being less than 2 is 0.9772. The final result is then<span><span>2(0.9772)−1=0.9544
Hope this is the answer that you are looking for.</span></span>
3 0
3 years ago
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