All categories

3 years ago

...................Help Please....

Mathematics

2 answers:

Contact [7]3 years ago

7 0

Answer:

Step-by-step explanation:

The slope is the ratio of the vertical difference between two points to the horizontal difference. This ratio is often referred to as "rise over run."

From the point (-2, -1) to the point (2, 7), the rise is 7-(-1) = 8, and the run is 2-(-2) = 4. The ratio of these numbers is ...

slope = rise/run = 8/4

slope = 2

_____

<em>Check</em>

The graph rises 2 units for a run of 1 unit, confirming this calculation.

gladu [14]3 years ago

5 0

Your answer should be 2.

You might be interested in

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

1 year ago

HELP PLEASE PLEASE HELP<br> Solve for t.<br> -10t − -17t − 10t + -13 = 11

Marta_Voda [28]

-10t - (-17t) - 10t + (-13) = 11 (two negative signs makes a positive)

-10t + 17t - 10t - 13 = 11 Combine like terms

-3t - 13 = 11 Add 13 on both sides

-3t = 24 Divide -3 on both sides

t = -8

8 0

3 years ago

Given h(x) = x^ - 2, what is the value of h(-3) ?<br><br> ^ = is exponent 2

ella [17]

Answer:

Step-by-step explanation:

<u>Given</u>

h(x) = x^ - 2
h(-3) = ?

<u>Substitute x with -3</u>

h(-3) = (-3)^-2 = 3^-2 = 9^-1 = 1/9

3 0

2 years ago

Read 2 more answers

13/15 is equal to or less than or greater than 0.5

Bess [88]

Answer:

Greater Than

Step-by-step explanation:

Half of 15 is 7 1/2 and 13 is greater than 7 1/2

5 0

1 year ago

m∠WYX=(2x−1)° and m∠WYZ=(4x+1)°. If ∠WYX and ∠WYZ are complementary, what is the measure of each angle?

Neko [114]

We are given : m∠WYX=(2x−1)° and m∠WYZ=(4x+1)°.

∠WYX and ∠WYZ are complementary.

We know, sum of complementary angles is = 90°.

So, we need to add ∠WYX and ∠WYZ and set it equal to 90°.

m∠WYX + m∠WYZ = 90°.

Plugging values of ∠WYX and ∠WYZ in the above equation, we get

(2x−1)° + (4x+1)° = 90°.

Removing parentheses from both sides,

2x-1 + 4x+1 =90.

Combining like terms,

2x+4x= 6x and -1+1 =0

6x +0 =90.

6x=90.

Dividing both sides by 6.

6x/6 =90/6

x= 15.

Plugging value of x=15.

m∠WYX=(2x−1)° = 2*15 -1 = 30 -1 =29

m∠WYZ=(4x+1)° = 4*15 +1 = 60+1 = 61.

Therefore, ∠WYX=29° and ∠WYZ=61°.

5 0

3 years ago