3/4 of an hour is 45 minutes. So they spent 3 hours and 45 minutes at the beach. They arrived at 10:30 so just add 3 hours and 45 minutes to their arrival time and you get 2:15pm. They left at 2:15pm
10:30+3:45=14:15
14:15=2:15
Car 1. (40 kph):
40 kph = 40,000 km per hour
Car 2. (15 mps)
15 mps x 60 (to get one minute) = 900 mpm
900 meter per minute
900 x 60 = 54,000 km
Therefore, car 2 is faster since it drives at 54km per hour and car 1 only travels at 40km per hour.
Also, you decide which car you want to choose.
Hope this helps! Have a great day.
Answer:
means that ab=c
Step-by-step explanation:
Answer:
see the explanation
Step-by-step explanation:
we know that
A shape with two opposite angles equal to 105° could be a quadrilateral, a parallelogram, a rhombus or a trapezoid
Because
<em>A quadrilateral</em>: A quadrilateral is a four-sided polygon. The sum of the interior angles in any quadrilateral must be equal to 360 degrees
so
If the quadrilateral have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to
<em>A parallelogram</em>: A Parallelogram is a flat shape with opposite sides parallel and equal in length. Opposite angles are congruent and consecutive angles are supplementary
so
If the parallelogram have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to
<em>A rhombus</em>: A Rhombus is a flat shape with 4 equal straight sides. A rhombus looks like a diamond. All sides have equal length. Opposite sides are parallel. Opposite angles are congruent and consecutive angles are supplementary
so
If the Rhombus have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to
<em>A trapezoid</em>: A trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides parallel
so
If the trapezoid have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to
I’m a little confused by the question and the problem, is there a picture of the problem you could add?