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2

Answer:

#(2x+5)(x+3)#

Explanation:

To factor this expression we will need to factor each of the terms individually so that when we reverse the factorization, the terms will re-multiply into the given expression.

Given: #2x^2 + 11x + 15#

Because we start out with #x^2# we know we will end up with two brackets: #( ...)( ...)#

We know that each bracket will need an #x# inside: #(x...)(x...)#

In this case we can see that one of the #x# terms will need to be multiplied by 2 since we start with #2x^2#. So #(2x...)(x...)#

We know that each bracket will need a numerical factor. In this case the number to be factored is: #15=(15)(1) =(3)(5)=(5)(3)#

So we can write in:#(2x...15)(x...1) = (2x...3)(x...5) = (2x...5)(x...3)#

But we need to have factors that when multiplied by #x# will add or subtract to result in the central term of the expression - in this case #11x#.

We can see that #2*3 = 6 and 5*1 = 5# will result in #11# when added. This indicates both numeric terms need to be #+#.

The given expression also agrees with the double #+#, because both signs are #+#.

Then we can write: #2x^2 + 11x + 15 = (2x+5)(x+3)#

To check for correctness, simply re-multiply the answer to result in the given expression.

1

Answer:

# vecr = ( (2), (1), (-1) )+ lamda ( (-14), (-9), (6) ) #

Explanation:

First consider the vectors #3hati-4hatj+hatk# and #2hatj+3hatk#. A vector #vecn#, perpendicular to these vectors is found by forming the cross product:

# vecn = ( (3), (-4), (1) ) xx ( (0), (2), (3) ) #

# \ \ \ \ = | (hati, hatj, hatk), (3,-4,1), (0,2,3) | #

# \ \ \ \ = | (-4,1), (2,3) | hati - | (3,1), (0,3) | hatj + | (3,-4), (0,2) | hatk#

# \ \ \ \ = (-12-2) hati - (9-0) hatj + (6-0) hatk#
# \ \ \ \ = -14 hati - 9 hatj + 6 hatk#

So now that we have the vector that the line is in the direction of, so the line equation is given by:

# vecr = vec(OG) + lamda vecr #
# \ \ \ \ = ( (2), (1), (-1) )+ lamda ( (-14), (-9), (6) ) #

We can confirm this with a 3D diagram:
enter image source here

1

Answer:

All the Kingdoms of of West Africa gained their wealth from the so called Gold salt trade with the Islamic states to the north.

Explanation:

Ghana was the first kingdom of west Africa. Using the technology of storing grain and iron working, the kingdom gained ascendancy in West Africa. Then with a monopoly with trade to the Islamic states became wealthy. The West African Kingdom would trade gold, ivory and slaves, for salt, manufactured goods, and cloth.

These precious goods from the north would then be traded with other tribes in Western Africa adding to the wealth. The Niger River was instrumental in moving goods within West Africa.

Ghana was replace by Mali as the dominate force in West Africa. Mali continued the same pattern of trade with the Islamic States. One of the kings of Mali Masa Musa ( meaning Great King) even traveled to Mecca bringing immense amounts of gold and slaves to the north

Mali was conquered by the Kingdom of Songhai. This kingdom followed the same pattern of trading gold, ivory and slaves with the north. The Niger River continued to be the main source of trade within Western Africa.

Then the Islamic Kingdom of Morocco decided to cut out the middle man. The Islamic Kingdom of Morocco destroyed the Songhai empire. However the other Western African kingdoms and tribes wouldn't trade with the Moroccans and the Gold Salt trade dissolved.

All three of the Western African kingdoms depended on the Gold Salt trade and control of the Niger River to dominate West Africa.

3

Yes, but not just one formula. There are many.

The derivative (of a differentiable function), #y=f(x)#, at the point #x=a# is defined by the following limit

# f'(a) = lim_(x rarr a )(f(x)-f(a))/(x-a) #

With a slight change of notation we can write:

# dy/dx = f'(x) = lim_(h rarr 0 ) (f(x+h)-f(x))/h #

In some older texts the notation may involve #deltax# or #Deltax#, instead of #h# giving an identical result:

# f'(x) = lim_(deltax rarr 0 ) (f(x+deltax)-f(x))/(deltax) = lim_(Deltaxrarr 0 ) (f(x+Deltax)-f(x))/(Deltax)#

It represents both the rate of change of the function, and the gradient of the tangent line at any particular point. If the limit does not exists then the function is not differentiable.

In practice we do not derive the derivative from first principles using the limit definition, but instead we use various rules that can be proved to be true;

Power Rule: #d/dx(x^n) = nx^(n - 1)#

Example: Find #dy/dx# for #y = 2x^7#.

We have #n = 7#, so the derivative is #dy/dx = 14x^(7-1) = 14x^6#

Chain Rule: #d/dx(f(g(x)) = f'(g(x)) * g'(x)#, in other words the derivative of the composition #f(g(x))# is the inner function times the outer function.

Example: Find #dy/dx# for #y = (x +2)^7#

We let #u = x + 2# and #y = u^7#. Then, by the power rule, #dy/(du) = 7u^6# and #(du)/dx = 1#.

#dy/dx = dy/(du) * (du)/dx#

#dy/dx = 1 * 7u^6#

#dy/dx = 7(x + 2)^6#

Product Rule: #d/dx(f(x) xx g(x)) = f'(x)g(x) + f(x)g'(x)#

Example: Find #dy/dx# for #y = (x + 2)(x^2 - 1)#

We let #f(x) =x + 2# and #g(x) = x^2 - 1#. Then #f'(x) = 1# and #g'(x) = 2x#.

#dy/dx = 1(x^2 - 1) + 2x(x + 2)#

#dy/dx = x^2 - 1 + 2x^2 + 4x#

#dy/dx = 3x^2 + 4x-1#

Quotient rule: #d/dx((f(x))/(g(x)))=(g(x)f'(x)-f(x)g'(x))/(g(x))^2#

Example: Find #dy/dx# for #y=x/(x + 1)#

We let #f(x)=x# and #g(x)=x + 1#. Then #f'(x)=1# and #g'(x)=1#
#dy/dx=(1(x + 1) - 1(x))/(x + 1)^2#

#dy/dx =1/(x + 1)^2#

Here are a few other useful derivative formulas I think you should know.

#d/dx(lnx) = 1/x#

#d/dx(sinx) = cosx#

#d/dx(cosx) = -sinx#

Once you get good at the basic differentiation rules, you may be asked to solve problems that combine the differentiation rules in interesting ways.

Example: Find #dy/dx# for #y = ln(secx)#

First of all, note that #secx = 1/cosx#. The function becomes

#y = ln(1/cosx)#

There are a few ways of differentiating this.

a) you could use the quotient rule to differentiate #1/cosx# and then the chain rule to differentiate #y#.

b) you could rewrite #1/cosx# as #(cosx)^-1# and then differentiate using the chain rule twice.

c) you could use the laws of logarithms to simplify and then differentiate. I'll use this method.

By the rule #ln(a/b) = lna - lnb#, we have:

#y = ln(1/cosx) = ln1 - lncosx = 0 - ln(cosx) = -ln(cosx)#

By the chain rule, we have

#dy/dx = -1/cosx * -sinx = sinx/cosx = tanx#

Hopefully you now get a good idea what differentiation is about!

1

Answer:

DNA replication happens in the nucleus of human cells. It will also take place within mitochondrial matrix.

Explanation:

In humans, DNA is found in the nucleus of cell. The process of replication (which copies DNA) must take place in the nucleus since this is where the DNA is found.

This video provides a brief summary of this process using the DNA Workshop activity from PBS.

DNA is also found in the mitochondria of human cells, so the process of replication will also take place in the mitochondria. Take that teacher who is trying to be tricky!!!

Hope this helps!

1

Answer:

#his# is a possessive pronoun.

Adverbs are used mostly to describe #verbs# which are the #action# in the sentence, but they can describe other components of the sentence structure.

Explanation:

#his# socks, #his# idea, #his# mother, all describe #objects# that belong to him. The pronoun is a result of referencing a subject indirectly. See below:

#His# dream in tatters, Neil decided to return to the drawing board.

The possessive pronoun #his# is followed immediately by the object #dream# and later the subject #Neil#.

2

Answer:

d) East

See below

Explanation:

The earth's magnetic field points South - North, ie the geographic North Pole is a south pole magnet-wise..... which is why the north pole of a compass points north.

We can now use the right hand rule to determine the direction of the force on a positively charged particle that is raining down on earth.

Problem is, there are a number of versions of the right hand (and left hand) rules.

I use this one because follows straight from the Lorentz Force Equation: #mathbf F = q (mathbf v times mathbf B)#:

http://montessorimuddle.org

So, using your right hand:

  • your forefinger points in the direction of travel of the positively charged particles (the old Benjamin Franklin idea of conventional current posits movement of positively charged particles). So you are pointing at the centre of the earth with that finger

  • your middle finger points in the direction of the B field, ie South - North

Your thumb should now be pointing left - right .... or West - East if you were in outer space with the geographic North Pole of the earth pointing vertically upwards. That is the direction of the force on the particles.

However, in the UK at the 15- 16 yo level, you have Fleming's right (generators) and Fleming's left (motors) hand rules. And the fingers are pointing at different things.

If we think of this as a motor, ie it's as if the earth-bound particles were a moving current within a conductor, intended to power motion of the conductor, then we use Fleming's left hand rule, and the fingers point as follows:

Wikipedia

We get the same result, of course:

  • the forefinger points South - North, same as the earth's B- field

  • the middle finger points at the centre of the earth, as that is the direction of travel of the positive charged particles

Once again, your thumb should now be pointing West - East.

2

Answer:

#Delta_"pH" = 0.299#

Explanation:

!! VERY LONG ANSWER !!

The first thing to do here is to figure out the concentrations of acetic acid and of acetate anions present in the buffer.

You know that a weak acid/conjugate base buffer has

#color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))) -># the Henderson - Hasselbalch equation

In your case, you have

#5.000 = 4.740 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

Even without doing any calculations, you should be able to say that this buffer contains more conjugate base than weak acid. This is the case because the pH of the buffer is greater than the #"p"K_a# of the acid.

You can rewrite the equation as

#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 5.000 - 4.740#

#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.26#

This will be equivalent to

#10^log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.26#

#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.8197#

This implies that

#["CH"_3"COO"^(-)] = 1.8197 * ["CH"_3"COOH"]#

You also know that the total molarity of weak acid and conjugate base is equal to #"0.100 M"#, so you can say that

#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.100 M"#

You now have two equations with two unknowns. For simplicity, I'll use

#["CH"_3"COOH"] = x" "# and #" "["CH"_3"COO"^(-)] = y#

You have

#x =0.100 - y#

Plug this into the fist equation to get

#y = 1.8197 * (0.100 - y)#

Rearrange to get

#2.8197 * y = 0.18197#

This will be equivalent to

#y = 0.18197/2.8197 = 0.06454#

This means that

#x = 0.100 - 0.06454 = 0.03546#

Therefore, you can say that the buffer contains

#["CH"_3"COOH"] = "0.03546 M"" "# and #" " ["CH"_3"COO"^(-)] = "0.06454 M"#

As predicted, the buffer contains more conjugate base than weak acid.

Now, you are adding

#7.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.360 moles H"_3"O"^(+))/(1color(red)(cancel(color(black)("L"))))="0.002520 moles H"_3"O"^(+)#

to the buffer by way of the hydrochloric acid solution. The hydronium cations will react with the acetate anions to produce acetic acid and water

#"CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> "CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l))#

Use the molarity of the acetate anions and the volume of the buffer to calculate the number of moles of conjugate base and of weak acid present in the buffer

#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.06454 moles CH"_3"COO"^(-))/(1color(red)(cancel(color(black)("L"))))#

# = "0.009681 moles CH"_3"COO"^(-)#

#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.03546 moles CH"_3"COOH")/(1color(red)(cancel(color(black)("L"))))#

# = "0.005319 moles CH"_3"COOH"#

The reaction will consume hydronium cations and acetate anions in a #1:1# mole ratio and produce acetic acid in a #1:1# mole ratio, so you can say that after the neutralization is complete, the solution will contain

#n_ ("H"_ 3"O"^(+)) = "0 moles" -># completely consumed

#n_ ("CH"_ 3"COO"^(-)) = "0.009681 moles" - "0.002520 moles"#

# = "0.007161 moles CH"_3"COO"^(-)#

#n_( "CH"_3 "COOH") = "0.005319 moles" + "0.002520 moles"#

# = "0.007839 moles CH"_3"COOH"#

The total volume of the solution will be

#V_"total" = 1.50 * 10^2"mL" + "7.00 mL" = "157 mL"#

The new concentrations of the weak acid and of the conjugate base will be

#["CH"_3"COOH"] = "0.007839 moles"/(157 * 10^(-3)"L") = "0.04993 M"#

#["CH"_3"COO"^(-)] = "0.007161 moles"/(157 * 10^(-3)"L") = "0.04561M"#

Finally, use the Henderson - Hasselbalch equation to calculate the new pH of the solution

#"pH" = 4.740 + log( (0.04561 color(red)(cancel(color(black)("M"))))/(0.04993color(red)(cancel(color(black)("M")))))#

#"pH" = 4.701#

Therefore, the pH of the solution changed by

#Delta_"pH" = |4.701 - 5.000| = color(darkgreen)(ul(color(black)(0.299)))#

The answer is rounded to three decimal places, the number of significant figures you have for your volumes and molarities.

So, does the result make sense?

You added hydrochloric acid, a strong acid, to the buffer, so you should expect its pH to decrease as a result.

Notice that the final concentration of the weak acid is greater than that of the weak base, which is why you have

#4.701 < 4.740" "# or #" ""pH" < "p"K_a#

1

Answer:

Euphony and cacophony both refer to a musical or rhythmic nuance applied to prose or poetry. They are opposites in their effect; euphony smooth and harmonious while cacophony is abrupt and jarring.

Explanation:

A silent breeze undulated the field of azure flax blooms imitating a rolling sea under the bright shine of the sun.

This is an example of euphony where the prose is written to inspire a calming effect on the reader.

We march, we drum, we thunder on; many more miles before we rest.

This is an example of cacophony in prose where the effect on the reader is more striking and upbeat.

1

Answer:

See explaination.

Explanation:

Fitzgerald, who was an expatriate himself, saw the hollowness of the poor. In this book, he shows the decline of the American Dream. There is no longer freedom, no individualism. Everything is greed, cheap thrills, and it's all money, all vulgar, all loneliness and empty souls through it all, with no happiness whatsoever.

Now to your question. There are two major things that cover the hollowness of the upper class:

The "valley of ashes" (the industrial dump) where part of the railroad runs through signifies the moral and social decay of America. It also signifies the remains of the persistent of money,

Another is the big billboard in the valley of ashes with Dr. Eckelburg's eyes. This signifies the eyes of God, looking over the country, and saddened by the lack of character.