12 ×4^4/4^2
=12×4^2
=12×16
=192
Answer:
The given point is a solution to the given system of inequalities.
Step-by-step explanation:
Again, we can substitute the coordinates of the given point into the system of inequalities. We know that the x-coordinate and y-coordinate of
are
and
, respectively.
Plugging these values into the first inequality,
, gives us
, which simplifies to
. This is a true statement, so the given point satisfies the first inequality. We still need to check if it satisfies the second inequality though, because if it doesn't, it won't be a solution to the system.
Plugging the coordinates into the second inequality,
, gives us
, which simplifies to
. This is also a true statement, so the given point satisfies the second inequality as well. Therefore,
is a solution to the given system of inequalities since it satisfies all of the inequalities in the system. Hope this helps!
Answer:
sin 2x + cos x = 2 sin x cos x + cos x = (2 sin x + 1)cos x
Step-by-step explanation:
Given the expression: sin 2x + cos x,
then we can use the formula: sin 2x = 2 sin x cos x, which gives:
sin 2x + cos x = 2 sin x cos x + cos x = (2 sin x + 1)cos x
So there you have two expressions in terms of sin x and cos x, as requested. :D
Answer: < -4/5, 3/5>
This is equivalent to writing < -0.8, 0.6 >
======================================================
Explanation:
Draw an xy grid and plot the point (-4,3) on it. Draw a segment from the origin to this point. Then draw a vertical line until reaching the x axis. See the diagram below.
We have a right triangle with legs of 4 and 3. The hypotenuse is
through use of the pythagorean theorem.
We have a 3-4-5 right triangle.
Therefore, the vector is 5 units long. This is the magnitude of the vector.
Divide each component by the magnitude so that the resulting vector is a unit vector pointing in this same direction.
Therefore, we go from < -4, 3 > to < -4/5, 3/5 >
This is equivalent to < -0.8, 0.6 > since -4/5 = -0.8 and 3/5 = 0.6
Side note: Unit vectors are useful in computer graphics.