Answer:
5
Explanation:
The sum of the digits of the number is ...
(4+1+3)+(4+6+5)+(7+8+9) = 8+15+24 = 47
The sum of those digits is 4+7=11, and those digits sum to 1+1 = 2.
That is, the value of the number mod 9 (or 3) is 2.
The ones digit is odd, so the value of the number mod 2 is 1.
This combination of modulo values tells you the mod 6 result is 5.
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<em>Additional comment</em>
We can look at the (mod2, mod3) values of the numbers 0 to 5:
0 ⇒ (0, 0)
1 ⇒ (1, 1)
2 ⇒ (0, 2)
3 ⇒ (1, 0)
4 ⇒ (0, 1)
5 ⇒ (1, 2) . . . . the mod {2, 3} results we have for the number of interest.
This process of adding up the digits repeatedly is referred to as "casting out 9s." The result of it is the modulo 9 value of the number (with 0 mapped to 9). Checking the mod 9 result of arithmetic operations is one quick way to spot certain kinds of errors. It can also be used as part of a divisibility test for 3 or 9.
Answer:
The main reason for this is the ingredients found in the paint. Different Ingredients are used to make different types of color paints, therefore when each of these paints is used in connection with the solvent the different chemicals in the paint create a certain reaction with the chemicals in the solvent, and each reaction may be different in each color depending on the chemical compound.
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
hello your question lacks the required image attached to this answer is the image required
answer : NOR1(q_) wave is complementary to NOR2(q)
Explanation:
Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_
Initial state is unknown i.e q = 0 and q_= 1
from the diagram the waveform reset and set
= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while
from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )
From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.
From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table
also from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.
since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)