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3 years ago

HELP ASAP how many solutions does the system of equations have? y=-2x+9 6x+3y+27

Mathematics

1 answer:

pav-90 [236]3 years ago

3 0

I believe the answer to your question is two

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eimsori [14]

Answer:

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Step-by-step explanation:

6 0

3 years ago

Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.

elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

$\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}$

Consider a vector field $F=x\^i+y\^j+z\^k$

Then the divergence of the vector F is,

div F = $\nabla.F$ = $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)$

and the curl of the vector F is,

curl F = $\nabla\times F$ = $\^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})$

<h3>Calculation:</h3>

The given vector fields are:

$F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$

1) Verifying the identity: $\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

Consider L.H.S

⇒ $\nabla.(aF1+bF2)$

⇒ $\nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))$

⇒ $\nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the dot product between these two vectors,

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(1)

Consider R.H.S

⇒ $a\nabla.F1+b\nabla.F2$

So,

$\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)$

⇒ $\nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}$

$\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)$

⇒ $\nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}$

Then,

$a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})$

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(2)

From (1) and (2),

$\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

2) Verifying the identity: $\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Consider L.H.S

⇒ $\nabla\times(aF1+bF2)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the cross product,

$\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})$ ...(3)

Consider R.H.S,

⇒ $a\nabla\times F1+b\nabla\times F2$

So,

$a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))$

⇒ $\^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})$

$a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))$

⇒ $\^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})$

Then,

$a\nabla\times F1+b\nabla\times F2$ =

...(4)

Thus, from (3) and (4),

$\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let $F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$ be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

$1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

8 0

2 years ago

Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles. Write an inequality that shows the distance johnathan

Mekhanik [1.2K]

<em><u>An inequality that shows the distance Johnathan could of ran any day this week is:</u></em>

$x\leq 3.5$

<em><u>Solution:</u></em>

Let "x" be the distance Johnathan can run any day of this week

Given that,

Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles

Therefore,

Number of days ran = 5

The most he ran in 1 day = 3.5 miles

Thus, the maximum distance he ran in a week is given as:

$distance = 5 \times 3.5 = 17.5$

The maximum distance he ran in a week is 17.5 miles

If we let x be the distance he can run any day of this week then, we get a inequality as:

$x\leq 3.5$

If we let y be the total distance he can travel in a week then, we may express it as,

$y\leq 17.5$

8 0

3 years ago

Marcus randomly draws tokens from a bag containing 10 blue tokens, 8 green tokens, and 12 red tokens. The first draw is a green

IrinaVladis [17]

Answer: 1. D) 30

2. A) 8

3. B) 4/15

4. A) 2/9

5.D) 64%

Step-by-step explanation:

1. Since, here the tokens from a bag containing 10 blue tokens, 8 green tokens, and 12 red tokens. The first draw is a green token.

Thus, the total token = 10 blue tokens+ 8 green tokens+ 12 red tokens = 30 token

Therefore, If the first draw is a green token, Then total outcomes that will possible = 30.

2. Since, getting green is an event with 8 favorable outcomes.

Therefore, If the first draw is a green token, Then total favorable outcomes that will possible = 8.

3. The possibility of getting green in first drawn, P(G) = $\frac{8_C_1}{30_C_1}$

P(G) = $\frac{8}{30} = \frac{4}{15}$

4. Here, Number of red socks= 5

Number of white socks = 2 and Number of blue socks=3

Therefore total socks = 10

⇒ Probability of picking a pair of red socks,

P(R)= $\frac{5_C_2}{10_C_2} = \frac{10}{45}$

⇒ P(R)= 2/9

5. Since, here coin is flipped 50 times and lands on heads 32 times.

Therefore, experimental probability that coin will land on heads on the next flip = (total times in which head appears/ total number of experiments) × 100 = (32/50)×100=64%

3 0

4 years ago

18x^2+24x=0 Please solve by factoring.

Harman [31]

Answer:

6x ( 3x +4)

X =0 and x = 4/3 or x = 1 1/3

Step-by-step explanation:

Take the common factors out,

Here x is common to both equation and also 6 is common to both equation

So,

6x ( 3x + 4) = 0

So it can be solved by

X = 0 and x = 4/3

4 0

4 years ago