identify the numerical coefficient (NC) and literal coefficient (LC) of each expression.1. 5 mNC:LC:2. 3+NC:LC:3. 10 a² b³NC:LC:4. -wNC:LC:5. -7stNC:LC:correct answer = 20 pointswrong answer = report
1. identify the numerical coefficient (NC) and literal coefficient (LC) of each expression.1. 5 mNC:LC:2. 3+NC:LC:3. 10 a² b³NC:LC:4. -wNC:LC:5. -7stNC:LC:correct answer = 20 pointswrong answer = report
identify the numerical coefficient (NC) and literal coefficient (LC) of each expression.
1. 5 m
NC: 5
LC: m
2. 3t
NC: 3
LC: t
3. 10 a² b³
NC: 10
LC: a²b³
4. -w
NC: -1
LC: w
5. -7st
NC: -7
LC: st
2. C.1.) 6C⁰ + 6C¹ + ... + 6C⁵ + 6C⁶2.) 7C⁰ + 7C¹ + ... + 7C⁶ + 7C⁷3.) 9C⁰ + 9C¹ + ... + 9C⁸ + 9C⁹4.) nC¹5.) mCm6.) nC³7. nC⁴8.) nC⁰D.1.) nC² = 212.)nC³ = 1653.) nC³ = 2n4.) (n - 2)C ⁿ-⁴ = 155.) nCⁿ-1 = 12
Answer:
C. Evaluating Combinations
Using the formula for the sum of combinations of n choose k with k from 0 to n:
nC0 + nC1 + ... + nCn = 2ⁿ
Substituting n=6, we get:
6C⁰ + 6C¹ + ... + 6C⁵ + 6C⁶ = 2⁶ = 64
Therefore, the value of the expression is 64.
Using the same formula as in (1):
7C⁰ + 7C¹ + ... + 7C⁶ + 7C⁷ = 2⁷ = 128
Therefore, the value of the expression is 128.
Using the same formula as in (1):
9C⁰ + 9C¹ + ... + 9C⁸ + 9C⁹ = 2⁹ = 512
Therefore, the value of the expression is 512.
nC1 = n
Therefore, the value of the expression is n.
mCm = 1
Therefore, the value of the expression is 1.
Using the formula for n choose k:
nCk = n! / (k! * (n-k)!)
Substituting k=3, we get:
nC³ = n! / (3! * (n-3)!)
Simplifying:
nC³ = n(n-1)(n-2) / 6
Therefore, the value of the expression is n(n-1)(n-2) / 6.
Using the formula for n choose k:
nCk = n! / (k! * (n-k)!)
Substituting k=4, we get:
nC⁴ = n! / (4! * (n-4)!)
Simplifying:
nC⁴ = n(n-1)(n-2)(n-3) / 24
Therefore, the value of the expression is n(n-1)(n-2)(n-3) / 24.
nC0 = 1
Therefore, the value of the expression is 1.
D. Solving Combination Equations
nC² = 21
Using the formula for n choose k:
nCk = n! / (k! * (n-k)!)
Substituting k=2, we get:
nC² = n! / (2! * (n-2)!)
Simplifying:
n(n-1) / 2 = 21
n² - n - 42 = 0
Factoring:
(n-7)(n+6) = 0
n=7 or n=-6
Since n has to be a positive integer, the solution is n=7.
Therefore, the value of n is 7.
nC³ = 165
Using the same formula as in (1) and substituting k=3, we get:
nC³ = n! / (3! * (n-3)!) = n(n-1)(n-2) / 6 = 165
Multiplying both sides by 6:
n(n-1)(n-2) = 990
Expanding and rearranging:
n³ - 3n² + 2n - 990 = 0
Factoring:
(n-10)(n+9)(n-11) = 0
n=10, n=-9, or n=11
Since n has to be a positive integer, the solution is n=11.
Therefore the value is 11
nC³ = 2n
Using the same formula as in (1) and substituting k=3, we get:
nC³ = n! / (3! * (n-3)!) = n(n-1)(n-2) / 6 = 2n
Multiplying both sides by 6:
n³ - 3n² + 2n = 12n
Rearranging:
n³ - 3n² - 10n = 0
Factoring out n:
n(n² - 3n - 10) = 0
Factoring the quadratic:
n(n-5)(n+2) = 0
n=0, n=5, or n=-2
Since n has to be a positive integer, the solution is n=5.
Therefore, the value of n is 5.
(n-2)Cⁿ⁻⁴ = 15
Using the formula for n choose k:
nCk = n! / (k! * (n-k)!)
Substituting k=n-4, we get:
(n-2)Cⁿ⁻⁴ = (n-2)! / ((n-4)! * 2!) = 15
Simplifying:
(n-2)(n-3)(n-4)(n-5) = 30
Expanding and rearranging:
n⁴ - 14n³ + 61n² - 92n + 60 = 0
We can try to find integer solutions by testing the factors of 60. Testing n=3, 4, and 5, we find that n=3 is a solution:
3⁴ - 14(3³) + 61(3²) - 92(3) + 60 = 81 - 126 + 183 - 276 + 60 = 0
Therefore, the value of n is 3.
nCⁿ-1 = 12
Using the formula for n choose k:
nCk = n! / (k! * (n-k)!)
Substituting k=n-1, we get:
nCⁿ-1 = n! / ((n-1)! * 1!) = n = 12
Therefore, the value of n is 12.
3. Q1 = 6.0 nC is at (0.30 m, 0); Q2 = -1.0 nC is at (0, 0.10 m); Q3 = 5.0 nC is at (0, 0). What is the direction of the net force on the 5.0 nC charge?
Answer:
56 above +x axis
Explanation:
4. who is ncs or nocoppyrightsounds??
Answer:
btw po salamat po talaga sa points sana po Hindi ninyo po Ako report pls po plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz
sorry po talaga
5. Find the missing value1. ⁶Cr=152. nC⁴=353. nC³=1204. ⁹Cr=1 5. ¹³Cr=715
Answer:
1. 4
2. 7
3. 10
4. 9
5.9
Step-by-step explanation:
I Use my scalculator with some brain.
6. A=(2,4,6,8) B=(2,4,10,12) (AuB)nc?
Answer:
(AuB)={2,4,6,8,19,12}
I don't know if I'm right so my apologies if I'm wrong
Step-by-step explanation:
I'm not that smart so I'm sorry if I'm wrong
7. 2. (BUD) nC solotion
Answer:
(BUD) nC = ( { 5, 7, 9 } u { 2,4,6,8,} ) n {1,3,5,7}
(BUD) nC = ( { 2,4,5,6,7,8,9} n { 1,3,5,7 }
(BUD) nC = { 5, 7 }
Step-by-step explanation:
{ 5,7,9} {2,4,6,8} {1,3,5,7} is the given
u is the union
n I forgot
8. what is the meaning of NC of smaw?
Answer:
Welding NC I
Explanation:
The Welding NC I (SMAW) Qualification consists of competencies that a person must achieve to weld carbon steel plates components as specified by layout, blueprints, diagrams, work order, welding procedure or oral instructions using SMAW welding equipment.
9. g_ _ d_ nc c_u_s_l_r
Answer:
↑
Explanation:
guidance counselor.
10. what is gelatin C or NC
Answer:
NC
Explanation:
I saw it at my book i just read it and saw that "NC" and i learned it and that's it
11. Find the electric at a point midway between two charges of +30.0 nC (left) and +60.0 nC (right) separated by a distance of 300 mm.
Answer:
To find the electric field at a point midway between two charges, we can use the principle of superposition, which states that the total electric field at a point due to multiple charges is the vector sum of the electric fields at that point due to each individual charge.
In this case, we have two charges of +30.0 nC and +60.0 nC, and we want to find the electric field at a point midway between them. We can use Coulomb's law to calculate the electric field due to each charge at the midpoint, and then add these fields vectorially to get the total electric field.
The distance from each charge to the midpoint is 150 mm (or 0.15 m), and the charges have the same sign, so they repel each other. Therefore, the electric field due to the +30.0 nC charge points to the left, and the electric field due to the +60.0 nC charge points to the right.
Using Coulomb's law, the magnitude of the electric field due to each charge at the midpoint is:
E = k * Q / r^2
where k is Coulomb's constant (8.99 x 10^9 N*m^2/C^2), Q is the charge in Coulombs, and r is the distance from the charge to the midpoint in meters.
For the +30.0 nC charge on the left:
E1 = k * Q1 / r^2
= (8.99 x 10^9 N*m^2/C^2) * (30.0 x 10^-9 C) / (0.15 m)^2
= 3.59 x 10^4 N/C to the left
For the +60.0 nC charge on the right:
E2 = k * Q2 / r^2
= (8.99 x 10^9 N*m^2/C^2) * (60.0 x 10^-9 C) / (0.15 m)^2
= 7.18 x 10^4 N/C to the right
The total electric field at the midpoint is the vector sum of these two fields:
E_total = E1 + E2
= 3.59 x 10^4 N/C to the left + 7.18 x 10^4 N/C to the right
= 3.59 x 10^4 N/C to the right
Therefore, the electric field at a point midway between a +30.0 nC charge and a +60.0 nC charge, separated by a distance of 300 mm, is 3.59 x 10^4 N/C to the right.
#RATEMEPLEASE
12. If q1=q4= 3.9x10 nC, q2=q5= -5.3 nC, and q3=q6=-3.9 nC. What is the net electric flux through the Gaussian surface? The value of the permittivity of free space is ε0= 8.85 x 10 -12 C2/ N.m2.
Answer:
The net electric flux through the Gaussian surface is - 598.87 N·m²/C.
Explanation:
Here, we are to solve for the total electric flux of three given charges in an enclosed Gaussian surface. In this problem, we will use the the concept of Gauss's law where the electric flux through any closed surface S is equal to the net charge enclosed divided by the permittivity of free space, given by:
So, what are we going to do is just add only the charges that are inside the enclosed surface. Out of the six charges, only three of them were inside the surface, so only these charges will be used to solve the electric flux.
For the given information
q1 = q4 = 3.9 nC
q2 = q5 = -5.3 nC
q3 = q6 = -3.9 nC
ε₀ = 8.85 x 10⁻¹² C²/ N.m²
Solving the problem
Let us use the formula for electric flux for any enclosed surface, we have:
[tex]\Phi=\frac{q_{enc}}{\epsilon_0 }[/tex]
Then substitute the given information, we have:
[tex]\Phi=\frac{3.9nC-5.3nC-3.9nC}{8.85 x 10^{-12}C^2/ N.m^2}[/tex]
3.9 nC will be cancelled because they are equal with opposite signs, so the remaining is
[tex]\Phi=\frac{-5.3nC}{8.85 x 10^{-12}C^2/ N.m^2}[/tex]
[tex]\Phi=\frac{-5.3x10^{-9}C}{8.85 x 10^{-12}C^2/ N.m^2}[/tex]
[tex]\Phi[/tex] = - 598.87 N·m²/C
Therefore, the net electric flux is - 598.87 N·m²/C.
To learn more, just click the link below:
Additional example with solution and answerbrainly.ph/question/2575789
#LetsStudy
13. Barayti nc wika halimbawa
Answer:
Mga Uri
May walong uri ng barayti ng wika: Idyotek, Dayalek, Sosyolek / Sosyalek, Etnolek, Ekolek, Pidgin, Creole, at Register.
1. Idyotek
Ito ay ang personal na paggamit ng salita ng isang indibidwal. Bawat indibidwal ay may istilo sa pamamahayag at pananalita.
Halimbawa:
“Magandang Gabi Bayan” – Noli de Castro
“Hoy Gising” – Ted Failon
“Hindi ka namin tatantanan” – Mike Enriquez
“Di umano’y -” – Jessica Soho
2. Dayalek
Ito ay nalilikha ng dahil sa heograpikonog kinaroroonan. Ang barayti na ito ay ginagamit ng mga tao ayon sa partikular na rehiyon o lalawigan na tnitirhan.
Halimbawa:
Tagalog – “Mahal kita”
Hiligaynon – “Langga ta gd ka”
Bikolano – “Namumutan ta ka”
Tagalog – “Hindi ko makaintindi”
Cebuano – “Dili ko sabot”
3. Sosyolek / Sosyalek
Uri ng barayti na pansamantala lang at ginagamit sa isang partikular na grupo.
Halimbawa:
Te meg, shat ta? (Pare, mag-inuman tayo)
Oh my God! It’s so mainit naman dito. (Naku, ang init naman dito!)
Wag kang snobber (Huwag kang maging suplado)
4. Etnolek
Ginawa ito mula sa salita ng mga etnolonggwistang grupo. Nagkaroon nga iba’t ibang etnolek dahil sa maraming mga pangkat na etniko.
14. Four point charges are equally separated by 1.00 m in air. If q1 = 3.0 nC, q2 = -4.0 nC, q3 = 2.0 nC, and q4 = -5.0 nC, what is the electric field at a point halfway between q2 and q3?
Answer:
1. REASONING AND SOLUTION The work done in moving a charge q0 from A to B is given by Equation 19.4: . For the cases in the drawing, we have
Case 1: 
Case 2: 
Case 3: 
The work done on the charge by the electric force is the same in each case.
_____________________________________________________________________________________________
2. REASONING AND SOLUTION The potential at a point in space that is a distance r from a point charge q is given by Equation 19.6: . When more than one point charge is present, the total potential at any location is the algebraic sum of the individual potentials created by each charge at that location.
A positive point charge and a negative point charge have equal magnitudes. One of the charges is fixed to one corner of a square. If the other charge is placed opposite to the first charge along the diagonal of the square, then each charge will be the same distance L from the empty corners. The potential at each of the empty corners will be


15. 333172152 ML I'd nc
salamat po sa points thank you po talaga laro din po tayou
16. give an example of NCS 1990
Answer:
early 1990's with a household sample of over 8000 respondents
Explanation:
i hope my answer can help you
module well!
17. Aclvity: Four point charges are equally separated by 1.0m in air. If q1=-3.0 nC, q2=-4.0 nC, q3=2.0.nC, and q4=-5.0 nC, what is the electric field at a point half way between q2 and q3?
To find the electric field at a point halfway between q2 and q3, we can use the principle of superposition. We'll break down the problem into finding the electric field at that point due to each individual charge, and then summing them up to find the total electric field.
The electric field due to a point charge at a distance r from the charge is given by:
E = kq/r^2
where k is Coulomb's constant, q is the charge, and r is the distance from the charge.
Let's label the point halfway between q2 and q3 as P. To find the distance from each of the charges to P, we can use the Pythagorean theorem. Since the charges are equally spaced, the distance from each charge to P is:
d = sqrt((1m/2)^2 + (1m/2)^2) = sqrt(1/2) m
Now we can calculate the electric field at P due to each charge:
Electric field at P due to q1:
E1 = kq1/d^2 = (9.0 x 10^9 N·m^2/C^2)(-3.0 x 10^-9 C)/(1/2)^2
E1 = -108 N/C
Electric field at P due to q2:
E2 = kq2/d^2 = (9.0 x 10^9 N·m^2/C^2)(-4.0 x 10^-9 C)/(1/2)^2
E2 = -144 N/C
Electric field at P due to q3:
E3 = kq3/d^2 = (9.0 x 10^9 N·m^2/C^2)(2.0 x 10^-9 C)/(1/2)^2
E3 = 36 N/C
Electric field at P due to q4:
E4 = kq4/d^2 = (9.0 x 10^9 N·m^2/C^2)(-5.0 x 10^-9 C)/(1/2)^2
E4 = -225 N/C
Now we can find the total electric field at P by summing up the electric fields due to each charge:
E_total = E1 + E2 + E3 + E4
E_total = -108 N/C - 144 N/C + 36 N/C - 225 N/C
E_total = -441 N/C
Therefore, the electric field at a point halfway between q2 and q3 is -441 N/C, directed from q4 to q1.
PA BRAINLLIEST PLEASE
18. ROELBONG18fhfhfngNCSo
ROEL
BONG18fhfhf
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NCS
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19. meaning of predicament NC or C
Answer:
it means no comment or comment
20. A – 41 nC charge is placed in between two-point charges, 32 nC and 24 nC. The negative charge moves from point a to b. What is the change in potential energy?
Answer:
pang grade 13 bato gawa po kayo bagong post
Explanation:
ung malinaw
21. Two small conducting and identical spheres A and B have charges -25 nC and 15 nC, respectively. They are separated by a distance of 0.02 m.
Answer:
2°nc and ~0°nc this are the separated
22. what is nc in tesda?
Answer:
National Certificate (NC)
Answer:
(National Certificate (NC)
23. Why does the Government requires everyone to take NC II or NC III?
Answer:
When you take a National Certificate from TESDA Assessment, you can choose from their program (depending on your chosen skills) to get more knowledge and experience related to it. If you have NC2, it will be easier for you to get the job you really wanted
24. what is the tolerance value for No color(NC)
Silver
hope this answer helped you have a great day!!
Answer:
SILVER
Explanation:
Mark as brainiest
25. what is the total electric force exerted by q1=4.65 nC at x=3 cm and q2= 5.61 nC at c=2 cm on a charge q3 =2.48 nC at x=1 cm?
Answer:
=1.12⋅10
−4N.
Explanation:
e total electric force on the third charge caused by charges 1 and 2 is the sum of the forces of both charges (1 and 2) acting on the charge 3 located at the beginning of coordinate (i.e. at x_3=0x
3
=0):
26. . Which is/are TRUE of a professional? –Analysis A. Completed college/university degree B. Required of NC IV from TESDA C. Demonstrates solely ethical competence D. Abides by his personal Code of Ethics
halamang gamot para sa pusong sinugatan
27. centrl nervous system (ncs)
Answer:
A nerve conduction study (NCS) --also called a nerve conduction velocity test (NCV)--is a measurement of the amount and speed of conduction of an electrical impulse through a nerve. NCS can determine nerve damage and destruction which cause the muscles to react in abnormal ways.
28. 3. A +25.0 nC charged object is 10.0 cm along a horizontal line toward the right of a -42.0 nC charged object as shown
Answer:
I'm sorry, but it is not clear to me what you are asking. Could you please provide more context or clarify your question?
29. nc g pre ︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎︎︎ ︎ ︎
Answer:
Hahah sayo victory para sa WR haha
Explanation:
Lakas mo mag mm
30. fraction to decimal bjcm NC d
Answer:
0.8
Explanation:
I think that's the answer
