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Fed [463]
3 years ago
12

Which of these equations are linear functions that go

Mathematics
1 answer:
serious [3.7K]3 years ago
8 0

Answer:

I don't think the answer exist

Step-by-step explanation:

for y=6x

when y=0, x=0 and when x=0, y=0

hence, (x,y)=(0,0)

for y=x+6

when y=0, x+6=0, x=-6 and when x=0, y=6

hence (x,y) = (-6,6)

for y=x^2+6

when x=0, y=6 and when y=0,

x^2+6=0

subtracting 6 from both sides

x^2=-6

take the square root of both sides and get

x=+ or -sqrt(-6)= + or - (2.45i)

i.e. x=+2.45i or x=-2.45i these are complex roots anyway, and its out of point.

for y=2x+6

when x=0, y=6 and when y=0,

2x+6=0,

2x=-6,

x=-3

the coordinates give (x,y)=(-3,6)

conclude....

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Someone please help me!!!!​
Liono4ka [1.6K]

Answer:   see proof below

<u>Step-by-step explanation:</u>

Use the Pythagorean Identity: cos²Ф + sin²Ф = 1

Use the Difference Identity: cos(A - B) = cosA · cosB + sinA · sinB

Use the Double Angle Identity: cos 2Ф = 1 - 2sin²Ф

<u>Proof LHS → RHS</u>

LHS:                                          (cosA - cosB)² + (sinA - sinB)²

Expand:          cos²A - 2cosA · cosB + <u>cos²B</u> + sin²A - 2sinA · sinB <u>+ sin²B</u>

Pythagorean Identity:   2 - 2cosA · cosB  - 2sinA · sinB

Factor:                           2(1 - (cosA · cosB  + sinA · sinB))

Difference Identity:       2(1 - (cos(A - B))

Let Ф = (A-B)/2:             2(1 - cos2Ф)

Double Angle Identity:  2(1 - (1 - 2sin²Ф))

Simplify:                         2(1 - 1 + 2sin²Ф)

                                   = 4sin²Ф

Substitute (A-B)/2=Ф:    4sin²(A-B)/2

LHS = RHS: 4sin²(A-B)/2 = 4sin²(A-B)/2   \checkmark

4 0
3 years ago
2 and 1/10 divided by 1 and 1/5
dexar [7]

Answer:

The answer is 1 and 3/4

Step-by-step explanation:


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Okay so, we divide 50 by 100. 50/100=0.5. Therefore, 50/100 as a decimal is 0.5 Hoped I helped. :)
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3 years ago
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