The specific volume of brass is 0.001865 ft3. what would be its density?
1 answer:
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Answer:
y = 1
d = 4
Step-by-step explanation:
8y - 2d = 0 ---------------(I)
5y + 4d = 26 1/4
5y + 4d = 105/4 ------------------(II)
(I)*2 16y - 4d = 0
(II) <u>5y + 4d = 105/4</u> {add (I) &(II)}
21y = 105/4
Substitute y =1 in equation (I)
8*1 - 2d = 0
8 - 2d = 0
-2d = -8
d = -8/-2
d = 4
Answer:
The wide of river is
Step-by-step explanation:
In this problem we know that
Triangles DEG and DEF are congruent by ASA (Angle-Side-Angle) Congruence
therefore
EG=EF