At the beginning of the season, MacDonald had to remove 5 orange trees from his farm. Each of the remaining trees produced 210 oranges for a total harvest of 41790 oranges.

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Answer

The given sequence is in the form of a Riemann sum for the integral of the function 1/(1+x) from 1 to 2. So, the correct answer is b) 'integral from 1 to 2 of (1/(x+1)) dx'.

The sequence in the question seems to be in the form of a Riemann sum for an integral. The term inside the summation loop can be expressed as 1/(1+i/n) where i varies from 1 to n. The limit of the sequence as n tends to infinity can be thus represented as the integral from 0 to 1 of the function 1/(1+x) dx. The function 1/(1+x) is a continuous function over the interval [0,1], so the integral exists and the limit of the sequence is legitimately defined.

Therefore, looking in the options provided, the answer to your question will be 'b.) integral from 1 to 2 of (1/(x+1)) dx'. This integral is the limit of the given series as n approaches infinity.

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The limit as 'n' approaches infinity of this summation is equal to the integral from 1 to 2 of (1/(x+1))dx. Essentially, this question is turning the sum into an integral as 'n' approaches infinity.

This question is asking for the limit of a sum as n approaches infinity, which is essentially the definition of an integral. Every term in the sum could be expressed as 1/(1+i/n), where 'i' is the sum index. Each term, when 'n' becomes infinitely large, becomes a term of the form i/n, or Δx, which is a small piece of the variable we’re integrating. Similarly, each 1/(1+i/n) term turns into 1/x for that small piece.

Given these transformations and the fact that the index i ranges from 1 to n, it's clear that as n approaches infinity, we are integrating the function 1/x from 1 to 2.

Thus, the correct answer would be: b.) integral from 1 to 2 of (1/(x+1))dx

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Answer:

56

Step-by-step explanation:

The answer is 56 just clarifying the top answer :))

The product of the given trinomials will have a degree of 4, and since there are 9 terms, the maximum possible number of terms for the product is 9.egree of 4 and a maximum possible number of terms of 9.

When multiplying two polynomials, the degree of the resulting polynomial is the sum of the degrees of the two polynomials being multiplied.

In this case, both trinomials have a degree of 2.

The maximum possible number of terms in the product of two polynomials is the product of the number of terms in each polynomial.

For the first trinomial, there are 3 terms, and for the second trinomial, there are also 3 terms.

Multiplying these together gives a maximum possible number of terms of 9.

To arrive at this conclusion, you can use the distributive property, where each term in the first trinomial is multiplied by each term in the second trinomial, resulting in a total of 9 possible term combinations.

Therefore, the product of the two trinomials will have a dTo multiply two trinomials, like [tex]\( (x^2 + x + 2)(x^2 - 2x + 3) \)[/tex], you apply the distributive property repeatedly, multiplying each term in the first trinomial by each term in the second trinomial and then summing up the products.

Each term in the first trinomial needs to be multiplied by each term in the second trinomial, resulting in a total of [tex]\( 3 \times 3 = 9 \)[/tex] products.

Now, regarding the degree, when you multiply two polynomials, you add the degrees of the polynomials.

Here, both polynomials are of degree 2, so the resulting polynomial will be of degree 2 + 2 = 4.

Thus, the product of the given trinomials will have a degree of 4, and since there are 9 terms, the maximum possible number of terms for the product is 9.egree of 4 and a maximum possible number of terms of 9.

Answer:

Amount of C-14 taken were 2.5 pounds and 3.5 pounds respectively.

Step-by-step explanation:

Radioactive decay is an exponential process represented by

[tex]A_{t}=A_{0}e^{-kt}[/tex]

where [tex]A_{t}[/tex] = Amount of the radioactive element after t years

[tex]A_{0}[/tex] = Initial amount

k = Decay constant

t = time in years

Half life period of Carbon-14 is 5730 years.

[tex]\frac{A_{0} }{2}=A_{0}e^{-5730k}[/tex]

[tex]\frac{1}{2}=e^{-5730k}[/tex]

Now we take ln (Natural log) on both the sides

[tex]ln(\frac{1}{2})=ln[e^{-5730k}][/tex]

-ln(2) = -5730kln(e)

0.69315 = 5730k

[tex]k=\frac{0.69315}{5730}[/tex]

[tex]k=1.21\times 10^{-4}[/tex]

Now we have to calculate the weight of samples of C-14 taken for the remaining quantities [tex]\frac{5}{8}[/tex] and [tex]\frac{7}{8}[/tex] of a pound.

[tex]\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{5}{8}=A_{0}e^{(-1.3863)}[/tex]

[tex]A_{0}=\frac{5}{8}\times e^{1.3863}[/tex]

[tex]A_{0}=\frac{5}{8}\times 4[/tex]

[tex]A_{0}=\frac{5}{2}[/tex]

[tex]A_{0}=2.5[/tex] pounds

Similarly for [tex]\frac{7}{8}[/tex] pounds

[tex]\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}[/tex]

[tex]\frac{7}{8}=A_{0}e^{(-1.3863)}[/tex]

[tex]A_{0}=\frac{7}{8}\times e^{(1.3863)}[/tex]

[tex]A_{0}=\frac{7}{8}\times 4[/tex]

[tex]A_{0}=\frac{7}{2}[/tex]

[tex]A_{0}=3.5[/tex] pounds

Answer:

D. LG = MN

Step-by-step explanation:

Final answer:

To prove two triangles congruent by SAS, we require equality of two sides and the included angle. Given line LT = line MQ and angle L = angle M, we also need line GT = line NQ and line LG = line MN. For the second part, line RS = line WU and line RT = line VU are necessary to fulfill SAS criteria.(Options C and D for first and Option E and F for second)

Explanation:

When proving two triangles are congruent using the Side-Angle-Side (SAS) postulate, you need to know that two sides and the included angle (the angle between the two sides) of one triangle are exactly equal to two sides and the included angle of another triangle.

In the given problem, we're told that line LT is congruent to line MQ and angle L is congruent to angle M. The additional information needed to prove triangle congruence by SAS would be to show that the second pair of sides around the included angle are also congruent (as in option C Line GT = line NQ) and to ensure congruence on the corresponding parts of the other triangle (as in option D Line LG = line MN).

Answering the second part, to prove congruence by SAS, we would need two corresponding sides and the included angle. Therefore, the correct choices are option E Line RS = line WU which provides the second side, and option F line RT = line VU ensures the sides are corresponding in the two triangles, enclosing the angle which is already given as equal.

Answer:

The factored form of the given expression is [tex]5a^2b^3(10b^2-7a^2+ab)[/tex]

Step-by-step explanation:

We have been given the expression [tex]50a^2b^5-35a^4b^3+5a^3b^4[/tex]

In order to factor it completely we can check for the GCF (greatest common factor) among all the three terms

The GCF is [tex]5a^2b^3[/tex]

On factor out the GCF, we are left with

[tex]5a^2b^3(10b^2-7a^2+ab)[/tex]

Therefore, the factored form of the given expression is [tex]5a^2b^3(10b^2-7a^2+ab)[/tex]

Answer:

given expression is [tex]5a^{2} b^{3} (10b^{2}-7a^{2} +ab)[/tex]

Step-by-step explanation:

Part A)

m = miles traveled

x = days on trip

y = reimbursement

55x + .45m = y

Part B)

55x + .45(300)= 2335

55x+ 135 = 2335

subtract 135 on both sides

55x= 2200

divide by 55 on both sides

x= 40

Part C) Rita Spent 40 Days on This Trip

Three numbers that round to 54.5 when rounded to the nearest tenth are 54.45, 54.55, and 54.55.

When rounding to the nearest tenth, we look at the digit in the hundredths place. If the digit is 5 or greater, we round up. If the digit is less than 5, we round down. In this case, we want to find three numbers that round to 54.5 when rounded to the nearest tenth. One possible set of numbers is 54.45, 54.55, and 54.55. When rounded to the nearest tenth, all of these numbers round to 54.5.

The situation represents the linear function.

Given that,

Based on the above information, we can say that

When dollar presents y value and the quarter presents x value

So, the slope should be [tex]\frac{1}{4}[/tex]

The linear function should be [tex]y = \frac{1}{4}x[/tex]

Therefore we can conclude that the situation represents the linear function.

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Answer:

A, B and E are correct.

Step-by-step explanation:

We are given a function representing the processing speed of computers with respect to time.

Let, x-axis = axis corresponding generation and y-axis = axis corresponding the processing speed.

We can see that the the starting point of the function is (0,16). This gives us that the range of the function is y≥16 and (0,16) represents the initial speed of the computer.

So, option A and B are correct.

As, we can see that the function is going upwards after passing the point (10,256). This gives us that, this point does not the represent the maximum processing speed.

So, option C is not correct.

Also, only the domain being x>0 does not imply that the processing speed increases.

So, option D is not correct

Further, as the graph is ascending, so we see that as time increases, the processing speed increases.

So, option E is correct.

Answer:

The correct statements are in bold.

A. The range is y≥16 , so the processing speeds are always positive. (As the initial point is 16 and the graph is going upwards from there hence,the range y≥16 shows positive speed growth. Below this range, the speed will fall.)

B. The point (0,16) represents an initial processing speed of 16 MHz. (This is clearly visible in the graph.)

C. is not right as the graph is going up from 10256.

D. This is also not right. The given domain is not specific of speed.

E. The graph is ascending, indicating that the processing speed increases over time. (We can see as the time increases, the processing speed increases.)

Answer:

Hence the orthocenter is (-2,12)

Step-by-step explanation:

We need to find the Orthocenter of ΔABC with vertices A(0,10) , B(4,10) and C(-2,4).

" Orthocenter of a triangle is a point of intersection, where three altitudes of a triangle connect ".

Step 1 :  Find the perpendicular slopes of any two sides of the triangle.  

Step 2 : Then by using point slope form, calculate the equation for those two altitudes with their respective coordinates.

Step 1 :  Given coordinates are: A(0,10) , B(4,10) and C(-2,4)

Slope of BC = [tex]\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}= \dfrac{4-10}{-2-4}=\dfrac{-6}{-6}=1[/tex]

Perpendicular Slope of BC = -1

( since for two perpendicular lines the slope is given as: [tex]m_{1}\times m_{2}=-1[/tex]

where  [tex]m_{1},m_{2}[/tex] are the slope of the two lines. )

Slope of AC = [tex]\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-10}{-2-0}=\dfrac{-6}{-2}=3[/tex]

Perpendicular Slope of AC= [tex]\dfrac{-1}{3}[/tex]

Step 2 :  Equation of AD, slope(m) = -1 and point A = (0,10)

[tex]y - y_{1} = m\times (x-x_{1})[/tex]

[tex]y - 10 = -1(x - 0)\\y - 10 = -x \\x + y = 10---------------(1)[/tex]

Equation of BE, slope(m) =\dfrac{-1}{3} and point B = (4,10)

[tex]y - y_1 = m(x - x_1)[/tex]

[tex]y - 10= \dfrac{-1}{3} \times (x - 4)[/tex]

[tex]3y-30=-x+4[/tex]

[tex]x+3y =34[/tex]----------(2)

Solving equations (1) and (2), we get

(x, y) = (-2,12)

Hence, the orthocenter is (-2,12).