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kobusy [5.1K]
2 years ago
15

3 sin A - cos A-2 sin A + 2 COSA​

Mathematics
1 answer:
Vlada [557]2 years ago
4 0

Answer:

The answer is sinA + cosA

Step-by-step explanation:

3 sinA – cosA – 2 sinA + 2 cosA

3 sinA – 2 sinA – cosA + 2 cosA

sinA + cosA

Thus, The answer is sinA + cosA

<u>-TheUnknownScientist</u><u> 72</u>

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So hmm, if you notice the picture below

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\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(10^o)=\cfrac{y}{2472}  solve for "y", to see how much north

and \bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies cos(10^o)=\cfrac{x}{2472}  solve for "x" to see how much west

make sure, that in both cases, your calculator is in Degree mode, since the angle is in degrees, as opposed to Radian mode


now... for the section B)

if notice the Reno-Miami line, and you were to draw a North line through it, anywhere in between, or at an endpoint like the in the picture at Reno, the "clockwise" angle, will always be the same, 100°

bear in mind ( no pun intended ), that Bearings use the North line, and the clockwise angle from it, as in the picture below

5 0
3 years ago
I've been stuck on this question for a while now, I need help!
MaRussiya [10]
In order to solve it you must make each component negative. Multiply them by negative one!

M < 2

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M is less than six. It could be 1, 2, 3, 4, or 5.

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Answer:

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Step-by-step explanation:

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Malini answered 68 problems on a test correctly and received a grade of 80%. How many problems were given on the test, if all pr
Levart [38]

Answer:

Below in bold.

Step-by-step explanation:

80% is equivalen to 68 correct answers

1%    ..        ..      ..       68/80      ..

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6 0
3 years ago
A.) Describe a transformation sequence that will transform quadrilateral ABCD into quadrilateral A’B’C’D’.
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a.) The orientation of ABCD is clockwise, as is the orientation of A'B'C'D'. This means the transformation involves a even number of reflections (may be 0). The orientation of AB is North, and the orientation of A'B' is West, so a rotation of 90° CCW (or equivalent) is involved. We can find the point of intersection of the perpendicular bisectors of AA' and BB' (at (-1, -1)) to determine a suitable center of rotation.

ABCD can be transformed to A'B'C'D' by ...

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8 0
3 years ago
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