Let the two numbers be x and y
x/y=3/4
Therefore x=3z
y=4z, where z is the GCF of x and y
(3z+k)/(4z+k)=4/5
(3z+k)*5=(4z+k)*4
15z+5k=16z+4k
k=z--->eq1
3z+k+4z+k=117
7z+2k=117--->eq2
Using value of z from eq1 in eq2
7k+2k=117
k=117/9
=13
Answer:
3 grams
Step-by-step explanation:
on the left side there is 3 circles and every circle is 3 grams, so there is 9 grams plus 6 squares for 2 grams its 12 . 9+12 =21 and two triangles with x mass
on the right side there is 2 circles , 2×3 = 6, and 3 squares, 3×2 = 6. 6+6=12 +5x
both sides are equal so we have
21+2x=12+5x
21-12=5x-2x
9=3x
x=3
Answer:
C 36
Step-by-step explanation:
Put numbers in order and count to the middle number