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V125BC [204]
2 years ago
13

25 students were surveyed about their favorite

Mathematics
1 answer:
Bess [88]2 years ago
5 0
Ik its messy but i hope this helps:]

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5+6+9•87•45•23-12-98•34/3
NISA [10]
I got 809293.333333, but the other dude might be right.
3 0
4 years ago
Bodmass<br><br>-3[2/6+8{-9/3(8-5*3)-6}]​
Vadim26 [7]

9514 1404 393

Answer:

  -361

Step-by-step explanation:

Your calculator can tell you the result. It is -361.

Start with the inner parentheses and work outward. Do multiplication and division in the order shown, left to right, before addition or subtraction.

  -3[2/6+8{-9/3(8-5*3)-6}]​

  = -3[2/6+8{-9/3(8-15)-6}]​

  = -3[2/6+8{-9/3(-7)-6}]​

  = -3[2/6+8{-3(-7)-6}]​

  = -3[2/6+8{21-6}]​

  = -3[2/6+8{15}]

  = -3[1/3+8{15}]

  = -3[1/3+120]

  = -3[361/3]

  = -361

8 0
3 years ago
Use Euler’s method to estimate the value of y(20). Use a step size of ∆x = 5. Please fill out the remainder of the chart complet
crimeas [40]

Answer:

y(20)=14,322

Step-by-step explanation:

We need to solve the equation:

dy/dx=0.4xy

with initial point

y(0) = 2

with incerments

\Delta x=h=5

By Euler's method, we have:

y(x+h)=y(x)+h*f'(x,y)

We start with the initial point

y(5)=y(0)+5*f'(0,2)=2+5*(0.4*0*2)=2

And we continue until we reach the value for y(20)

y(10)=y(5)+5*f'(5,2)=2+5*(0.4*5*2)=2+20=22\\\\\\y(15)=y(10)+5*f'(10,22)=22+5*(0.4*10*22)=22+440=462\\\\\\y(20)=y(15)+5*f'(15,462)=462+5*(0.4*15*462)=462+13,860\\\\y(20)=14,322

4 0
3 years ago
The length of rectangle ABCD is 4 in. The length of similar rectangle DEFG is 6 in. How many times greater than the area of ABCD
Fiesta28 [93]
The length of the DEFG rectangle
is 6/4 of the length of the ABCD rectangle.

6/4=3/2 = 1,5

How many times greater than the surface of ABCD is the DEFG area?
1,5 times
7 0
4 years ago
Use a Strategy to find the sum of 29-32
Lana71 [14]
29 - 32 = -3.  so yeahhhh
7 0
4 years ago
Read 2 more answers
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