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3 years ago

Help please, would be greatful!

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

7 0

Step-by-step explanation:

Work out the squares by themselves, if each square is 2, to work out the area=base x height

if each scale has been enlarged by 2 then 2x2=4 each square is 4cm², there are 5 squares

4x5=20cm²

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The length of a rectangle is two more than twice the width. the perimeter is 58ft. what is the area of the rectangle?

goldenfox [79]

58/2 = 29

w +2w+2 = 29

3w+2 =29

3w = 27

w = 27/3 = 9

width = 9

length = 2*9=18+2 = 20

20+20+9+9 = 58

Length = 20 feet

width = 9 feet

5 0

3 years ago

the probability tree diagram shows the probabilities that a football team will win their first two games find the probability of

nevsk [136]

Answer:1/2

Step-by-step explanation:

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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

Greta currently works 45 hours per week and earns a weekly salary of $729. She

marissa [1.9K]

Answer:

Greta will be earning 3.6/h more at her new job.

Step-by-step explanation:

To find the difference subract Greta‘s salary from her current job from Greta’s salary from her future job.

To find Greta’s salary (numerator) at her future job divide her current weekly salary ($729) by 45h then multiply that answer (16.2) by the increase in salary 10% or 1.1. You should end up with $17.82- this is your numerator. To find the denominator divide her current hours (45h) by 45 then multiply that answer (1) by 0.9 which is the 10% decrease in hours. Now you have a fraction ($17.82/0.9) then multiply that fraction by 1.1/1.1 to get the denominator to 1 hour so you can subtract the fractions. You should end up with $19.8/h.

Now subtract Greta’s current salary ($16.2/h)- you just take the numbers from the first part before you increase or decrease- from her future salary ($19.8/h), you will end up with 3.6/h.

I was challenged to write this in a single equation:

[((($729\45h)x1.1)/((45h\45)x0.9))x1.1/1.1]-[($729\45h)/(45h\45)]

=$3.6/h

/ means a fraction bar

\ means division

Also I am just a student so please tell me if you find any mistakes I could fix or any suggestions to make this a better explanation, and if you have any questions ask away.

4 0

3 years ago

Find the horizontal asymptote

eimsori [14]

Answer:

Correct answer: h = 3/5 or y = 3/5

Step-by-step explanation:

Given:

h(x) = (3x² - 9x - 4) / (5x² - 4x + 8)

h = lim x -> ∞ (3x² - 9x - 4) / (5x² - 4x + 8) the numerator and denominator will be divided by x² and get

h = lim x -> ∞ (3 - 9(1/x) - 4 (1/x²)) / (5 - 4(1/x) + 8 (1/x²)

The terms 1/x and 1/x² when x strives ∞ they strives 0

x -> ∞ ⇒ 1/x -> 0 and x -> ∞ ⇒ 1/x² -> 0

h = (3 - 0 - 0) / (5 - 0 + 0)

Horizontal asymptote is h = 3/5 or y = 3/5

God is with you!!!

5 0

4 years ago