Answer:
J'(-2,5)
K'(-7,5)
M'(-3.5,2)
L'(5.5,2)
Step-by-step explanation:
Please refer to the image attached to this answer.
Trapezoid JKLM is reflected over the x-axis, what do we notice about the ordered pairs of the original figure and the ordered pairs of its reflection over the x-axis is described below.
Point J , K L and M , when reflected over x axis , there distance from the y axis remains the same in second quadrant. the x coordinate becomes negative however the y coordinates remains the same.
The coordinates will be as shown in the image attached.
Answer: D
Step-by-step explanation: By using SOH for sin A, 'S' being sin, 'O' being opposite side of angle A and 'H' being the hypotenuse which is the longest part of the triangle you would find that 15 is opposite from Sin A and 17 the hypotenuse, 15/17.
For cos A you would use CAH, C= cos, A which is the adjacent of the triangle located next to angel A which is 8, and H= hypotenuse (also note that the hypotenuse never changes even if the angle may be different) CAH would be cos A = 8/17
Answer:
E. 64,000,000
Step-by-step explanation:
2^26 = (2^13)^2
But 2^13 = 8000 so
2^26 = (8,000)^2
= 64,000,000
9514 1404 393
Answer:
C. Obtuse
Step-by-step explanation:
The "form factor" I use for this is ...
a^2 + b^2 - c^2 = 6^2 + 8^2 - 11^2 = 36 +64 -121 = -21
The value is negative, indicating an OBTUSE triangle.
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<em>Additional comment</em>
In this expression, 'a' and 'b' are the two shortest sides (in no particular order) and 'c' is the longest side. The interpretation is ...
negative — obtuse
zero — right
positive — acute
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If you're familiar with Pythagorean triples, you know that the 3-4-5 right triangle triple can be doubled to give 6-8-10. The long side of 11 is longer than the hypotenuse for the right triangle, so would correspond to a largest angle greater than 90°. The 6-8-11 triangle would be OBTUSE.