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3 years ago

What is the perimeter, P, of a rectangle that has a length of x + 6 and a width of y - 1?

Mathematics

2 answers:

Mekhanik [1.2K]3 years ago

5 0

Answer: P = 2x + 2y + 10

Step-by-step explanation: The perimeter (P) of a rectangle is given as

P = 2L + 2W

By factorizing the right hand side of the equation we now have

P = 2(L + W)

However, the length of the rectangle is given as x + 6 and the width is given as y - 1

If P = 2(L + W), then

P = 2(x + 6 + {y - 1} )

P = 2(x + 6 + y -1)

P = 2(x + y + 5)

After expanding the bracket, we now have

P = 2x + 2y + 10

Gnom [1K]3 years ago

3 0

Answer:

just took the test the answer is P = 2x + 2y + 10

Step-by-step explanation:

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Please help ASAP. Trigonometry

Contact [7]

Answer:

cos(52°) = 18/x

x = 18·sec(52°)

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you that ...

Cos = Adjacent/Hypotenuse

In this geometry, that means ...

cos(52°) = 18/x

You can use the relation sec(x) = 1/cos(x) to rewrite this as ...

x = 18·sec(52°)

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You can also use the complementary angle and the complementary trig function.

sin(90° -52°) = 18/x

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2 years ago

A coin is tossed 20 times. It lands on heads 9 times. What is the P(H) according to your experiment? Write your answer as a frac

Mama L [17]

Answer:

$\frac{9}{20}=0.45=45$ %

Step-by-step explanation:

Probability is found by first forming a fraction where the numerator represents the number of desired outcomes and the denominator represents the total number of outcomes. In this case, the desired outcomes is the amount of times the coin lands on heads (9) and the total outcomes is the number of times the coin is tossed (20):

$\frac{9}{20}$

Divide numerator by denominator to get a decimal: 9÷20= 0.45

Multiply the decimal by 100 to get a percent: 0.45 x 100 = 45%

8 0

4 years ago

Last Question...Please help if you can...(25 points if you answer this)

tresset_1 [31]

Answer:

Annie- 17 hours, Brian- 3 hours

Step-by-step explanation:

Let the number of hours Annie travelled be a, and same of Brian be b

a + b = 20
a = 5b + 2

<u>Substituting a in the first equation</u>

5b + 2 + b = 20
6b = 20 - 2
6b = 18
b = 3

<u>Then finding a</u>

a = 20 - 3 = 17

<u>So the answer is:</u>

Annie- 17 hours, Brian- 3 hours

5 0

4 years ago

Olivia picked 482 apples from the orchard’s largest apple tree. She divided the apples evenly into 4 baskets. How many apples ar

Verdich [7]

Answer:

120 with 2 left over

Step-by-step explanation:

482/4 equals 120.5 so each basket gets 120 and then .5+.5+.5+.5= 2 whole extra apples

7 0

3 years ago

Read 2 more answers

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

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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.

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

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2 years ago