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tresset_1 [31]
2 years ago
13

Please help me!! This is due at 11:59 pm

Mathematics
1 answer:
scoray [572]2 years ago
3 0
Pretty sure it’s -2/1
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Please help I need a answer on this question
mestny [16]
Your answer would be D.....

Why well because rising bike is green which is more time and more hrs, playing video games would be purple and green would be t.v so yeah

3 0
2 years ago
Choose the function whose graph is given by:
wolverine [178]

<em>Greetings from Brasil...</em>

In a trigonometric function

F(X) = ±UD ± A.COS(Px + LR)

UD - move the graph to Up or Down (+ = up | - = down)

A - amplitude

P - period (period = 2π/P)

LR - move the graph to Left or Right (+ = left | - = right)

So:

A) F(X) = COS(X + 1)

standard cosine graph with 1 unit shift to the left

B) F(X) = COS(X) - 1 = -1 + COS(X)

standard cosine graph with 1 unit down

C) F(X) = COS(X - 1)

standard cosine graph with shift 1 unit to the right

D) F(X) = SEN(X - 1)

standard Sine graph with shift 1 unit to the right

Observing the graph we notice the sine function shifted 1 unit to the right, then

<h3>Option D</h3>

<em>(cosine star the curve in X and Y = zero. sine start the curve in Y = 1)</em>

7 0
2 years ago
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A student solved this problem and said the answer was 8 1/4 feet. Jenny had three pieces of rope. One piece was 3 1/2 ft long, a
Zigmanuir [339]
The answer would be B hope I helped u out!!!!!☺☺☺☺☺☺
3 0
2 years ago
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How do you subtract fractions with different denominators?
PilotLPTM [1.2K]
Attached the solution and work.

6 0
3 years ago
Does this graph represent a function ? Why or why not ?
balu736 [363]

Answer:

D) It passes the horizontal line test

Step-by-step explanation:

In order for a line to represent a function it must pass the vertical line test.

To apply the horizontal line test, take a pencil and hold it vertically. Then, move it across the graph from left to right. If it intersects the line at 1 point everywhere along the line, then it is a function (functions can only have 1 Y value for every X value).

3 0
3 years ago
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