12 red counters
7 more red counters than yellow counters
12 - 7 = 5 yellow counters
Dara has 5 yellow counters.
If you are given with all the tree sides of the triangle, you may solve for all the angles through the Law of Cosines,
c² = a² + b² - 2ab(cos C)
where angle C is the angle opposite the side c. You may use the same equation to get the values of the remaining angle. Additionally, if you already have one known angle, you can solve for the rest of the angles by Law of Sines,
a / sin A = b / sin B = c / sin C
Answer:
You will need 20 sides to complete the loop.
Step-by-step explanation:
The question isn't quite clear given how small the corner is, but I assume that we are looking to complete the circle if the pentagon and square are repeated in a loop
We can also see - assuming that those are proper equal-sided polygons, that PQ is the same length as PV
With that in mind, We can solve this by noting that the angle of a corner in a square is 90 degrees, and in a pentagon it's 108 degrees.
108 - 90 is equal to 18. This means that PQ is at eighteen degrees to YP. Also, QM, (which will be equivalent to the next VP is eighteen degrees to PQ.
This means that each polygon is rotated 18 degrees relative to it's neighbour.
With all that we can say that the total polygons we need to form a circle is 360/18 = 20, So you will need 20 polygons, or ten squares and ten pentagons to complete the loop.
Answer:
I'm newbie here
Step-by-step explanation:
Elimination Method.
4x + 6/y = 15 → step 1
6x - 8/y = 14 → step 2
so,
4x + 6/y = 15 |×6|
6x - 8/y = 14 |×4|
24x + 6/y(6) = 90
24x - 8/y(4) = 56
24x + 36/y = 90
24x - 32/y = 56
____________ _
68/y = 34
68 = 34y
34y = 68
y = 2
subsitution y = 2 to..
4x + 6/y = 15
4x + 6/2 = 15
4x + 3 = 15
4x = 12
x = 3
So, for x is 3, and for y is 2
