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BaLLatris [955]
2 years ago
10

Brainliest for help! Sketch a graph of the function y =sin(x) + 2

Mathematics
1 answer:
diamong [38]2 years ago
4 0

Answer:

The parent sine graph y=sin(x) has a range of -1 ≤ y ≤ 1

It crosses the x-axis x = 0° ± 180°n

The maximum points occur when x = 90° ± 360°n and y = 1

The minimum points occur when x = 270° ± 360°n and y = -1

The sketch the graph of function y=sin(x)+2 we simply move the graph of y=sin(x) up 2 units.

So this means it will have a range of 1 ≤ y ≤ 3

It no longer crosses the x-axis.

The maximum points occur when x = 90° ± 360°n and y = 3

The minimum points occur when x = 270° ± 360°n and y = 1

<u>Attached diagram</u>

The parent function y=sin(x) is shown in grey (dashed line)

The function y=sin(x)+2 is shown in red.

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anyanavicka [17]

Step-by-step explanation:

about circle R

arc angle MP = angle MRP = 78°

angle M = angle LMP.

according to the rules of inscribed angles, LMP is half of angle LRP.

LRP + MRP = 180° because together they cover the whole LM segment (half-circle).

LRP = 180 - MRP = 180 - 78 = 102°

so, LMP = angle M = 102/2 = 61°

also circle R (with arc ULNP)

the arc ULNP = arc UL + arc LN + arc NP = 160°

due to the congruent definition of JU and MP we know that arc UL = arc NP.

and we know arc LN = arc JM = 18°.

so,

160 = 2×arc NP + 18

142 = 2×arc NP

arc NP = 71° = arc UL

now, because arc MP + arc LP = 180 (half-circle),

arc MP = 180 - arc LP = 180 - arc LN - arc NP =

= 180 - 18 - 71 = 91°

about the circle R with 2 congruent chords :

the second answer : they are equidistant from R (the center of the circle).

about circle W

arc MK + arc KL = 180° (half-circle).

and we see

arc HL = arc KL = 53°.

so,

arc MK + 53 = 180°

arc MK = 180 - 53 = 127°

about circle M

arc QTS = 204°.

therefore, arc QRS = 360 - arc QTS = 360 - 204 =

= 156° = angle QMS.

according to the rules of inscribed angles

angle QTS = QMS/2 = 156/2 = 78°

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