Hello from MrBillDoesMath!
Answer:
Your answer is correct. (You have earned the unofficial title of Genius, Jr. );
Discussion:
Whew! This is a long problem. Let me see what I get and compare it to your answer. First each of the 3 terms contains the sqrt(yz) so let's factor that terms out. You equations equals
sqrt(yz) ( sqrt(108) + 3 sqrt(98) + 2 sqrt(75) ) (A)
Hmm.... let's look at the prime factorizations....
108 = 3 * 36 = 3 * (4*9) = 3 * 2^2* 3^2
98 = 2 * 49 = 2 * 7^2
75 = 3 * 25 = 3 * 5^2
So (A) =
sqrt(yz) ( 6 sqrt(3) + 3*7 sqrt(2) + 2*5 sqrt(3) ) =
sqrt(yz) ( ( 6 + 2*5) sqrt(3) + 21 sqrt(2) ) =
sqrt(yz) ( (6 + 10) sqrt(3) + 21 sqrt(2)) =
sqrt(yz) ( 16 sqrt(3) + 21 sqrt(2) )
Our answers agree!
Thank you,
MrB
Answer:
300
Step-by-step explanation:
Let x represent number of girls.
18 girls / 30 students
x girls / 500 students
18/30 = x/500
x = 18/30 x 500
x= 300 girls
Therefore there are approximately 300 girls in the school.
Step-by-step explanation:
after seeing your method of your addition I can answer it like 69....
hope it helps....
Answer:
62-12=50
Step-by-step explanation:
hope that helps!
Answer:
.
See the diagram attached below.
Let the chords be AB and AC with common point A.
AD is the diameter. Join B with D and C with D to form two triangles.
We need to prove that AB=AC.
\begin{gathered}In\ \triangle ABD\ and \triangle ACD;\\Given\ that\ \angle BAD=\angle CAD----(condition\ 1)\\since\ AD\ is\ diameter, \angle ABD=\angle ACD = 90^0\\So\ \angle ADB=\angle ADC--------(condition\ 2)\\AD=AD\ (common\ side)-----(condition\ 3)\\ \\So\ the\ triangles\ are\ congruent\ by\ ASA\ rule.\\Hence\ AB=AC.\end{gathered}
In △ABD and△ACD;
Given that ∠BAD=∠CAD−−−−(condition 1)
since AD is diameter,∠ABD=∠ACD=90
0
So ∠ADB=∠ADC−−−−−−−−(condition 2)
AD=AD (common side)−−−−−(condition 3)
So the triangles are congruent by ASA rule.
Hence AB=AC.
