Criminal evidence is any physical or verbal evidence presented in court to prove a specific crime. Evidence can take different forms and can be used by the accused to prove innocence. Admissible evidence, moreover, is the use of testimony (oral or written such as affidavit), exhibits (physical objects), documentary material or demonstrative evidence ( in the form of representation of an object including photos, x-rays, videotapes, movies, sound recordings, diagrams, forensic animation, maps, drawings, graphs, animation, simulations or models). All are presented before a judge or jury to prove a point or element in a case. In the choices above, six of them can qualify as individual evidence except numbers 3. Number 8, however, is questionable since it has incomplete description as to where and how is the paint was chipped.
Answer:
for movement of water through an organism
Explanation:
transpiration is the process by which water is transported from the roots to the leaves of a plant through the xylem
Answer: Abiotic factors in a tropical rainforest include temperature, humidity, soil composition, air, and many others. A few of the many biotic factors in a rainforest would include toucans, frogs, snakes, and anteaters.
Explanation: abiotic factors are non-living things. biotic factors are living things. they both impact the organism living in the environment. remember all biotic factors are dependent among abiotic factors.
Answer:
The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.
Explanation:
From the given information:
For food source A; we have:
3P₁ + P₂ + 2P₃ = 58 units of food A ---- (1)
For food source B; we have:
2P₁ + 4P₂ + 2P₃ = 70 units of food B ---- (2)
For food source C; we have:
P₁ + P₂ = 20 units of food C ----- (3)
From equation (1) and (2); we have:
3P₁ + P₂ + 2P₃ = 58
2P₁ + 4P₂ + 2P₃ = 70
By elimination method
3P₁ + P₂ + 2P₃ = 58
-
2P₁ + 4P₂ + 2P₃ = 70
<u> </u>
<u> P₁ - 3P₂ + 0 = - 12 </u>
P₁ = -12 + 3P₂ ---- (4)
Replace, the value of P₁ in (4) into equation (3)
P₁ + P₂ = 20
-12 + 3P₂ + P₂ = 20
4P₂ = 20 + 12
4P₂ = 32
P₂ = 32/4
P₂ = 8
From equation (3) again;
P₁ + P₂ = 20
P₁ + 8 = 20
P₁ = 20 - 8
P₁ = 12
To find P₃; replace the value of P₁ and P₂ into (1)
3P₁ + P₂ + 2P₃ = 58
3(12) + 8 + 2P₃ = 58
36 + 8 + 2P₃ = 58
2P₃ = 58 - 36 -8
2P₃ = 14
P₃ = 14/2
P₃ = 7
Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.
