Plz help! Student A
Draw a line that intersects points B and C.
Draw a line through point and point G.
Keep the compass at the same width, and place it on point C.
Keeping the compass at the same width, place it on point F.
Mark the intersection of the two arcs as point G.
Open the compass to the width between points D and E.
Place the compass on point B, and swing an arc that crosses line AB and
line BC. Label the points D and E.
Swing an arc that crosses line BC, and label the point F.
Swing an arc that intersects the arc created from line BC at point C.
Student B
Draw a line that intersects points B and C.
Place the compass on point B, and swing an arc that crosses line AB
and line BC. Label the points D and E.
Keep the compass at the same width, and place it on point C.
Swing an arc that crosses line BC, and label the point F.
Open the compass to the width between points D and E.
Keeping the compass at the same width, place it on point F.
Swing an arc that intersects the arc created from line BC at point C.
Mark the intersection of the two arcs as point G.
Draw a line through point C and point G.

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step-by-step explanation:

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answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕

29933333.33

step-by-step explanation:

here we can see that the storage increases

at certain times of the year - during the

warmer months, when the gas is not much

used for heating and reserves can be

increased - and then decreases during the

colder months because the supply is being

used. it is possible to model this with a

relatively simple trigonometric graph, but

first we examine the simplest trig curves.

065/065.html

the basics

the simplest sine and cosine curves can be generated by plotting the trig-function values

against the angles used in each function. the values can be found in the chart below:

ø (˚) 0 30 45 60 90 120 135 150 180 210 225 240 270 300 315 330 360

sin ø

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0 1

2

2

2 3

2 1 3

2 2

2 1

2

0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0

cos ø

€

1 3

2 2

2 1

2 0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0 1

2 2

2 3

2 1

first we plot the sine data, letting the x-axis represent the angles and the y-axis the sine

values. the result is the scatter plot below.

based on just these 17 data points, we can see

that the function is following a straightforward

pattern - the sine values are increasing to 1,

decreasing through the same values to -1, then

returning to zero. when we connect the dots, we

get the periodic curve y = sin (x) below.

this “sine curve,” with equation

y = sin(x), is the simplest of trig

graphs. the curve rises and falls by

one unit from its “axis of

oscillation” - in this case, the x-axis

- and after we cycle through 360˚ it

appears we are ready to go through

the same y-values again. we can see

this same pattern in the cosine

function on the following page.

each sine and cosine curve has two basic characteristics:

amplitude - distance the curve rises above, and falls below, its axis of oscillation (marked by

the vertical arrows on the cosine graph). the formula is

€

amp = max− min

2 .

period - the length of one cycle (marked by the horizontal arrows on the cosine graph).

this can be calculated by measuring the distance between any successive maximum

points (or minimum points), or by doubling the horizontal distance between a

minimum and the next maximum.

for our first two curves, then, the amplitude is 1 and the period is 360˚ (or 2! if we are in radians).

adjusting the amplitude and frequency: y = a sin(bx), y = a cos(bx)

now that we have our “base” amplitude and frequency . . let’s change them. we can adjust both

characteristics by introducing the parameters a and b into the equations. the graph below shows the

curves y = a sin(x) for a = 1,2,3, and -0.5, with the original curve y = sin(x) being the thickest line.

clearly the amplitude of the curves has changed;

when a = 2, the amplitude is 2; when a = 3, the

amplitude is 3; and when a = -0.5 the amplitude is

0.5 (since amplitude, a distance, must be positive).

however, this last curve has also been flipped upside

down (or “reflected in the x-axis”). since the

amplitude must be positive even when a itself is

negative, we have the formula

amplitude = | a |.

the relationship between the b-value and the

period is not so straightforward. on the right

we have y = cos(bx) for b = 1,3, and 0.5. (the

thickest line is again our “base” curve.) when

b=3, the period decreases to 120˚; when

b=0.5, the period increases to 720˚ (so only

half the curve is shown here). we can describe

this reciprocal relationship with the formula

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b = 360

period or

€

b = 2π

period ,

using the latter if we are working in radians.

shifting horizontally and vertically: y = sin(x-h) + k, y = cos(x-h) + k

to shift (or “translate”) a curve along either the x- or y-axis (or both), we introduce the parameters

h and k. below we have the graph of y = sin (x - 30) + 2 and y = sin(x). (note that we have returned

to an amplitude of 1 and period of 360˚). our original

curve y = sin(x) had its peak at the point (90,1); the

shifted curve has its corresponding peak at (120, 3). in

the same way, the original curve had a minimum value

at (270, -1); now the minimum is at (300,1). and

finally, the axis of oscillation has shifted vertically

from the line x = 0 to the line x =2. in other words, our

new curve has been created by shifting y = sin(x) right

30 and u p 2. and, in general, we can create the

following shifts:

horizontal shift = h (left if h < 0, right if h > 0)

vertical shift/new axis of oscillation = k.

modelling data: y = a sin[b(x-h)] + k, y = a cos[b(x-h)] + k

we have seen how the basic sine and cosine curves can be adjusted: by changing the amplitude

and period, and by shifting the curves horizontally and vertically. a summary is below.

amplitude = | a | vertical shift/axis of oscillation = k

period =

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360

b

or

€

2π

b