Answer:
- The area of rhombus JKLM is 48 units²
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<h3>Given </h3>
- Rhombus JKLM,
- Vertices at J(-2, -4), K(2, 2), L(6, -4), M(2, -10).
<h3>To find </h3>
<h3>Solution</h3>
We know that diagonals of rhombus are perpendicular to each other.
Hence its area is half the product of diagonals.
The diagonals are JL and KM and one of them is vertical and the other one horizontal since x- or y-coordinates are equal in pairs.
Let's find the length of diagonals, using the difference of coordinates:
- JL = 6 - (-2) = 8 units,
- KM = 2 - (-10) = 12 units.
Now find the area:
- A = JL*KM /2 = 8*12 / 2 = 48 units²
Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
Answer:
It would be D the angles are unrelated
Step-by-step explanation:
It would be D because PTQ and QTS doesn't look alike with angles