Answer: Hot air from the sand rose and was replaced by cool air blowing in from the ocean as a sea breeze.
Step-by-step explanation:
sorry it's late but hopefully it will help others
Answer:
C. ∠SRT≅∠VTR and ∠STR≅∠VRT
Step-by-step explanation:
Given:
Quadrilateral is a parallelogram.
RS║VT; RT is an transversal line;
Hence By alternate interior angle property;
∠SRT≅∠VTR
∠STR≅∠VRT
Now in Δ VRT and Δ STR
∠SRT≅∠VTR (from above)
segment RT= Segment RT (common Segment for both triangles)
∠STR≅∠VRT (from above)
Now by ASA theorem;
Δ VRT ≅ Δ STR
Hence the answer is C. ∠SRT≅∠VTR and ∠STR≅∠VRT
<span>Semicircle area = circle area / 2 = (pi*r^2)/2
16 in is the radius. Let's call the area of this semicircle s.
3.14*16^2=s
</span><span>Trapezoid area = ((base 1 + base 2) / 2) * h
Here, base 1 is 44 and base 2 is 36. The height (h) is 20. Let's call this trapezoid's area t.
(44+36)/2 *20 = t
</span>
<span>We can remove the area of the semicircle (that created by the windshield wiper) by subtracting it from the area of the trapezoid.
t - s = uncleaned area
[(44+36)/2 * 20] - [3.14 * 16^2] = uncleaned area</span>
Answer:
65°
Step-by-step explanation:
Radii CA and CB are perpendicular to tangent lines AT and BT, so
Since angle BAT is equal to 65°, angle CAB has measure
Consider triangle ACB. This triangle is isosceles, because CA=CB as radii of the circle. Two angles adjacent to the base are congruent, thus
The sum of the measures of all interior angles in triangle is always 180°, so
Angle ACB is central angle subtended on the minor arc AB, angle APB is inscribed angle subtended on the same minor arc AB. The measure of inscribed angle is half the measure of central angle subtended on the same arc, so
