first, consider that 25 = 5^2, and -25 = -1*25 = -1*5^2.
Then, you can write:
because i = √-1.
Hence, the answer is 5i
*100 points* What is the area of the parallelogram?
2 answers:
Answer:
12
Step-by-step explanation:
'To determine the area of the parallelogram, you need the value of h so that you can compute A = b * h
sin(A) = opposite / hypotenuse = h / a
This means that you can compute h like this:
h = sin(A) * a
Your area is...
A = b * h = sin(A) * a * h'
Now, you can fill in the numbers for the variables and result to your answer.
Another short way to find you answer is to use a very similar equation...
A = (h) * (a) * sin(A)
A = (4) * (6) * sin(60°)
A = 24sin(60°)
A = 24 * ( / 2)
A = 12
You might be interested in
Explanation:
When the inequality symbol is replaced by an equal sign, the resulting linear equation is the boundary of the solution space of the inequality. Whether that boundary is included in the solution region or not depends on the inequality symbol.
The boundary line is included if the symbol includes the "or equal to" condition (≤ or ≥). An included boundary line is graphed as a solid line.
When the inequality symbol does not include the "or equal to" condition (< or >), the boundary line is not included in the solution space, and it is graphed as a dashed line.
Once the boundary line is graphed, the half-plane that makes up the solution space is shaded. The shaded half-plane will be to the right or above the boundary line if the inequality can be structured to be of one of these forms:
- x > ... or x ≥ ... ⇒ shading is to the right of the boundary
- y > ... or y ≥ ... ⇒ shading is above the boundary
Otherwise, the shaded solution space will be below or to the left of the boundary line.
_____
Just as a system of linear equations may have no solution, so that may be the case for inequalities. If the boundary lines are parallel and the solution spaces do not overlap, then there is no solution.
_____
The attached graph shows an example of graphed inequalities. The solutions for this system are in the doubly-shaded area to the left of the point where the lines intersect. We have purposely shown both kinds of inequalities (one "or equal to" and one not) with shading both above and below the boundary lines.
