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DochEvi [55]
3 years ago
7

What are some equivalent fractions for 1/20

Mathematics
2 answers:
Talja [164]3 years ago
3 0
Some equivalent fractions of 1/20 are:
<span>1/20 = 2/40 = 3/60 = 4/80 = 5/100 = 6/120 = 7/140 = 8/160 = 9/180 = 10/200 = 11/220 = 12/240 = 13/260 = 14/280 = 15/300 = 16/320 = 17/340 = 18/360 = 19/380 = 20/400</span>
galben [10]3 years ago
3 0
Some fractions are 2/20, 3/60, 4/80, and 5/100.
To find equivalent fractions you can multiply the numerator and denominator by any number but has to be the same number
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Slope-intercept form<br> find out the y intercept and slope
Alex17521 [72]

Answer:

18. Slope (m) = 1/2

Y-intercept (b) = -6

19. Slope (m) = 2

Y-intercep (b) = 8

20. y = 1/2 + 6

Step-by-step explanation:

y = mx + b

y - 2x = 8

y = 2x + 8

7 0
3 years ago
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Consider the sequence of numbers: 3/8, 3/4, 1 1/8, 1 1/2, 1 7/8. . . Which statement is a description of the sequence?
AlekseyPX

Answer:

The sequence is recursive and can be represented by the function f(n + 1)= f(n) + 3/8

Step-by-step explanation:

i got it right on the assignment

7 0
3 years ago
What is the solution to the equation 3(2x+5)=3x+4x?
lina2011 [118]

Answer:

x=15

Step-by-step explanation:

6x+15=7x

15=x

7 0
3 years ago
P= (x,y) is an arbitrary point on the circle x^2+y^2=36
sesenic [268]

Let (x,y) be an arbitrary point on the circle.

Then d = √[x-3)2 + (y-0)2] = √[x2 - 6x + 9 + y2]

Since x2 + y2 = 1, y2 = 1-x2.

So, d = √[x2 - 6x + 9 + 1 - x2]

     d = √(-6x+10)

Domain:  x is the x-coordinate of a point on the circle centered at (0,0) with radius 1.  So, -1≤x≤1.

             But, for d to be defined, we need -6x+10 ≥ 0.  So, x ≤ 5/3  (True for all x in [-1,1]).

             Domain = [-1,1]

Range:    -1 ≤ x ≤ 1   So, 6 ≥ -6x ≥ -6

                                    16 ≥ -6x+10 ≥ 4

                                     4 ≥ √(-6x+10) ≥ 2   That is, 2 ≤ d ≤4

              Range:  [2,4]

Step-by-step explanation:

3 0
3 years ago
Find the area between y = 8 sin ( x ) y=8sin⁡(x) and y = 8 cos ( x ) y=8cos⁡(x) over the interval [ 0 , π ] . [0,π]. (Use decima
Marina86 [1]

Answer:

0.416 au

Step-by-step explanation:

Let y1=8sin(x) and y2=8cos(x), we must find the area between y1 and y2

\int\limits^\pi _0{(8cos(x)-8sin(x))} \, dx = 8\int\limits^\pi _0{(cos(x)-sin(x))} \, dx =\\8(sin(x)+cos(x)) evaluated(0-\pi )=\\8(sin(\pi )-sin(0))+8(cos(\pi )-cos(0))=\\8(0.054-0)+8(0.998-1)=8(0.054)+8(-0.002)=0.432-0.016=0.416

3 0
3 years ago
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