Can someone help me with this question please
1 answer:
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Answer:
The length of each of the other two sides would be 3.1 cm
Step-by-step explanation:
<u><em>The options of the question are</em></u>
A. 3.1 cm
B. 3.2 cm
C. 4.2 cm
D. 6.1 cm
we know that
An isosceles triangle has two equal sides and two equal interior angles
Let
x -----> the length of each of the two equal sides in the isosceles triangle
The perimeter of one isosceles triangle is equal to
Remember that
To maximize the use of the wire, the perimeter of 6 earrings (3 pairs of earrings) must be equal to 50 cm
so
Solve for x
therefore
The length of each of the other two sides would be 3.1 cm
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.
