First we have to assume that each quarter touched each other. Hence the area of the table not covered by the coins (A) is equal to the total area of the table (At) minus the total area of the coins (Ac). Coins are circle, so
and r =24.26mm. The area of one coin is then 1848.98mm^2. Hence the equation is A = At - xAc where x is the number of coins.
Answer:
(6,0) I think
my memory is rusty on this subject sorry if it's wrong.
Answer:
∠MSR is right angle
Step-by-step explanation:
It is given that RHOM is a Rhombus
Also ΔHOM, ΔMHR, ΔRHO and ΔOMR are isosceles triangles
Let us take ΔMSR and ΔRSH
∠MRS =∠HRS ( since it is given that the diagonal RO bisects ∠R)
∠RMS =∠RHS ( since Δ MRH is isosceles triangle )
RS = RS ( common side )
By AAS congruency rule ΔMSR ≅ ΔHSR
so we have
∠MSR=∠RSH ( corresponding parts of congruent triangles are congruent)
also we have
∠MSR +∠RSH =180° ( supplementary angles)
∠MSR +∠MSR=180° ( since ∠MSR=∠RSH)
2∠MSR= 180°
∠MSR =90°
Hence ∠MSR is right angle
Answer:
One solution, A) (2 , -1)
Step-by-step explanation:
The solution is where the two lines intersect. If the lines do not intersect, it is no solution. If the line intersects once (like in the picture), it has one solution. If it intersects more than once (x), then it will have x amount of solutions.
In this case, the intersection is at (2 , -1), therefore, A) (2 , -1) is your answer.
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