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OleMash [197]
3 years ago
15

Can someone help me please??

Mathematics
2 answers:
Viktor [21]3 years ago
6 0

Answer:

D

Step-by-step explanation:

Since you start out with an initial amount and are adding each time, it would increase with a positive slope, left to right

Sedaia [141]3 years ago
6 0

Answer:

D. The graph will be increasing from left to right.

Step-by-step explanation:

Both of the numbers given are positive, so the graph must also be positive.

If this answer is correct, please make me Brainliest!

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The set of data in the table below represents a linear function.
Vinil7 [7]

Answer:

y = 2x - 19

Step-by-step explanation:

Graph the points on the table and you'll be able to see the answer. That's how I went about it.

or use m = y^2 - y^1 divided by x^2 - x^1 to find the slope, then solve for the initial value from there

8 0
1 year ago
Reduce 30/36 to its most simple form
Novay_Z [31]
5/6 is the answer. Hope this helps you!
3 0
3 years ago
Read 2 more answers
Can someone help me with the question? Thanks! <br> xoxo
horrorfan [7]
2. because each x number times 2 plus one is equal to the y range
7 0
3 years ago
Answer the question in the picture
Svet_ta [14]
<h3>Answer:</h3>
  • left picture (bottom expression): -cot(x)
  • right picture (top expression): tan(x)
<h3>Step-by-step explanation:</h3>

A graphing calculator can show you a graph of each expression, which you can compare to the offered choices.

_____

You can make use of the relations ...

... sin(a)+sin(b) = 2sin((a+b)/2)cos((a-b)/2)

... cos(a)+cos(b) = 2cos((a+b)/2)cos((a-b)/2)

... cos(a)-cos(b) = -2sin((a+b)/2)sin((a-b)/2)

Then you have ...

\dfrac{\cos{2x}-\cos{4x}}{\sin{2x}+\sin{4x}}=\dfrac{2\sin{3x}\sin{x}}{2\sin{3x}\cos{x}}=\dfrac{\sin{x}}{\cos{x}}=\tan{x}

and ...

\dfrac{\cos{2x}+\cos{4x}}{\sin{2x}-\sin{4x}}=\dfrac{2\cos{3x}\cos{x}}{-2\cos{3x}\sin{x}}=\dfrac{-\cos{x}}{\sin{x}}=-\cot{x}


6 0
4 years ago
(Please 75 points if answerd need this is 1 min)
chubhunter [2.5K]

Answer:

(5,-4)

Step-by-step explanation:

If reflected over the x-axis, the quadrilateral would be in the fourth quadrant. N' would be at (1,-1) and P' at (6,-1). To reflect, look at the y-coordinate of the point and turn it to negative. With point Q, it's at (5,4) so we just flip the 4 to -4 and that's our point! (5,-4)

8 0
2 years ago
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