lim
x →1+. 1- x/x² - 1

If [tex]a[/tex] is the first term and [tex]r[/tex] is the common ratio of the series, then the sum of the first [tex]n[/tex] terms is

[tex]S_n = a + ar + ar^2 + \cdots + ar^n[/tex]

Multiply by [tex]r[/tex] on both sides, then subtract that from and solve for [tex]S_n[/tex].

[tex]rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n+1}[/tex]

[tex](1-r)S_n = a - ar^{n+1}[/tex]

[tex]S_n = \dfrac{a(1-r^{n+1})}{1-r}[/tex]

Use the given first and second terms of the sequence to find [tex]a[/tex] and [tex]r[/tex].

[tex]a = \dfrac1{64}[/tex]

[tex]ar = \dfrac1{32} \implies \dfrac r{64} = \dfrac1{32} \implies r = \dfrac{64}{32}=2[/tex]

Solve for [tex]n[/tex].

[tex]S_n = \dfrac{1-2^{n+1}}{64(1-2)} = 2^{36} - 2^{-6}[/tex]

[tex]\dfrac{2^{n+1} - 1}{2^6} = 2^{36} - 2^{-6}[/tex]

[tex]2^{n-5} - 2^{-6} = 2^{36} - 2^{-6}[/tex]

[tex]\implies n-5 = 36 \implies \boxed{n = 41}[/tex]

The speed of an object is the relationship between the distance covered by the object and the time taken. Thus, the average speed of the other car is 30 mph.

The speed of an object is the relationship between the distance covered by the object and the time taken. The expression for speed is given as;

speed = [tex]\frac{distance covered}{time taken}[/tex]

⇒ distance = speed x time taken

In the given question, the distance of the first car after 2 hours can be determined as:

distance = 80 mph x 2 h

              = 160 miles

The distance covered by the first car after 2 hours is 160 miles.

Thus, since the total distance between the two cars after 2 hours is 220 miles, then;

distance covered by the second car = 220 miles - 160 miles

                                                          = 60 miles

The distance covered by the second car is 60 miles.

So that the average speed of the other car is;

speed = [tex]\frac{60}{2}[/tex]

           = 30 mph

The average speed is 30 miles per hour (mph).

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Answer:

the third answer (-3, -3, 3)

Step-by-step explanation:

the dotted lines indicate that these lines (or part of lines) are invisible due to being inside or behind the cube.

so, we know that A is the left-back-upper corner of the cube.

the x-axis goes front-back. front is positive.

the y-axis goes left-right. right is positive.

the z-axis goes up-down. up is positive.

since the origin (0, 0, 0) is the center of the cube, and the sides are all 6 units long, the corners can have only -3 or +3 coordinates (half of a side length on either side of the origin) in each direction.

as we said : left-back-upper corner.

left means negative y.

back means negative x.

upper means positive z.

so, the coordinates of A are

(-3, -3, 3)

The width and height of the rectangle inscribed in the right triangle have a measure of 3.529 units.

In this problem we must use critical values and algebraic methods to determine to determine the dimensions of the rectangle such that the area is a maximum. The equation of the quadrilateral is formed by definition of the area of a rectangle:

A = w · h    (1)

Where:

And the area of the entire triangle is:

0.5 · (5) · (12) = w · h + 0.5 · w · (12 - h) + 0.5 · (5 - w) · h

30 = w · h + 6 · w - 0.5 · w · h + 2.5 · h - 0.5 · w · h

30 = 6 · w + 2.5 · h

2.5 · h = 30 - 6 · w

h = 12 - 2.4 · w    (2)

The quadrilateral of maximum area is always a square, then we must  solve for w = h:

w = 12 - 2.4 · w

3.4 · w = 12

w = 3.529

Then, the width and height of the rectangle inscribed in the right triangle have a measure of 3.529 units.

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The time needed for the investment to become $10,413 is 54.7 years

Number of years = (In FV / PV) / r  

r = 4.252/12 = 0.354%

(In 10, 413 / 1,018.40) / 0.00354 = 656.73 months = 54.7 years

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Let [tex]S(t)[/tex] denote the amount of salt (in lbs) in the tank at time [tex]t[/tex] min up to the 10th minute. The tank starts with 100 gal of fresh water, so [tex]S(0)=0[/tex].

Salt flows into the tank at a rate of

[tex]\left(0.5\dfrac{\rm lb}{\rm gal}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = 1\dfrac{\rm lb}{\rm min}[/tex]

and flows out with rate

[tex]\left(\dfrac{S(t)\,\rm lb}{100\,\mathrm{gal} + \left(2\frac{\rm gal}{\rm min} - 2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{S(t)}{50} \dfrac{\rm lb}{\rm min}[/tex]

Then the net rate of change in the salt content of the mixture is governed by the linear differential equation

[tex]\dfrac{dS}{dt} = 1 - \dfrac S{50}[/tex]

Solving with an integrating factor, we have

[tex]\dfrac{dS}{dt} + \dfrac S{50} = 1[/tex]

[tex]\dfrac{dS}{dt} e^{t/50}+ \dfrac1{50}Se^{t/50} = e^{t/50}[/tex]

[tex]\dfrac{d}{dt} \left(S e^{t/50}\right) = e^{t/50}[/tex]

By the fundamental theorem of calculus, integrating both sides yields

[tex]\displaystyle S e^{t/50} = Se^{t/50}\bigg|_{t=0} + \int_0^t e^{u/50}\, du[/tex]

[tex]S e^{t/50} = S(0) + 50(e^{t/50} - 1)[/tex]

[tex]S = 50 - 50e^{-t/50}[/tex]

After 10 min, the tank contains

[tex]S(10) = 50 - 50e^{-10/50} = 50 \dfrac{e^{1/5}-1}{e^{1/5}} \approx 9.063 \,\rm lb[/tex]

of salt.

Now let [tex]\hat S(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex] min after the first 10 minutes have elapsed, with initial value [tex]\hat S(0)=S(10)[/tex].

Fresh water is poured into the tank, so there is no salt inflow. The salt that remains in the tank flows out at a rate of

[tex]\left(\dfrac{\hat S(t)\,\rm lb}{100\,\mathrm{gal}+\left(2\frac{\rm gal}{\rm min}-2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{\hat S(t)}{50} \dfrac{\rm lb}{\rm min}[/tex]

so that [tex]\hat S[/tex] is given by the differential equation

[tex]\dfrac{d\hat S}{dt} = -\dfrac{\hat S}{50}[/tex]

We solve this equation in exactly the same way.

[tex]\dfrac{d\hat S}{dt} + \dfrac{\hat S}{50} = 0[/tex]

[tex]\dfrac{d\hat S}{dt} e^{t/50} + \dfrac1{50}\hat S e^{t/50} = 0[/tex]

[tex]\dfrac{d}{dt} \left(\hat S e^{t/50}\right) = 0[/tex]

[tex]\hat S e^{t/50} = \hat S(0)[/tex]

[tex]\hat S = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-t/50}[/tex]

After another 10 min, the tank has

[tex]\hat S(10) = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-1/5} = 50 \dfrac{e^{1/5}-1}{e^{2/5}} \approx \boxed{7.421}[/tex]

lb of salt.

Answer:

26.556192

Step-by-step explanation:

7.2×10460÷100

20000÷(7.2×10460÷100)

We need 56 pints of the 55 % pure fruit juice and 84 pints of the 80 % pure fruit juice to prepare 140 pints of 70 % pure fruit juice.

In this problem we have to use two kinds of juice with different concentrations. To get a certain concentration, we must adjust quantities of each kind and weighted averages offers a method that is easy to apply:

70 · 140 = 55 · x + 80 · (140 - x)

70 · 140 = 80 · 140 - 25 · x

25 · x = 10 · 140

x = 56

We need 56 pints of the 55 % pure fruit juice and 84 pints of the 80 % pure fruit juice to prepare 140 pints of 70 % pure fruit juice.

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Answer:

2.5

Step-by-step explanation:

2.5 is the only positive number amongst the options, so it must be the greatest.

The linear equations are given as follows:

A linear function is modeled by:

y = mx + b

In which:

For diagonal AC, the line contains points (-3,3) and (3,-3), hence the line is:

y = -x.

In standard form:

[tex]x + y = 0, -3 \leq x \leq 3[/tex].

For diagonal BD, the line contains points (3,3) and (-3,-3), hence the line is:

y = x.

In standard form:

[tex]x - y = 0, -3 \leq x \leq 3[/tex].

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Answer:

B

Step-by-step explanation:

Hope this helps:)

Answer:

B & D

Step-by-step explanation:

A The variable "x" is canceled  does not apply

B     One solution

C                                                  does not apply

D      One solution

Hope this helps

Answer:

B, C

Step-by-step explanation:

Look at the lower blue point.

Its coordinates are (4, 3).

The x-coordinate, 4, stands for minutes.

The y-coordinate, 3, stands for number of boxes.

This is a slope of 3/4, corresponding to 0.75 boxes per minute, or 3 boxes every 4 minutes.

Answer: B, C

Answer:

3 boxes are filled every 4 minutes.

Step-by-step explanation:

Look at the graph; at the bottom of the graph, it is shown that every __ minutes, __ boxes are filled. If you look at the first shown blue point on the left of the graph, it shows that on the fourth minute, three boxes are filled, as from the axis, you go right four and up three.

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

Area left for flowers = Area of flower garden - Area of bird bath.

Area of flower garden [ rectangle ] :

[tex]\qquad❖ \: \sf \:length \times width[/tex]

[tex]\qquad❖ \: \sf \:3 \times 2[/tex]

[tex]\qquad❖ \: \sf \: 6\: \: {m}^{2} [/tex]

Area of bird bath [ Circle ] :

[tex]\qquad❖ \: \sf \: \pi{r}^{2} [/tex]

[tex]\qquad❖ \: \sf \:3.14 \times ( \frac{1}{2} ) {}^{2} [/tex]

[tex]\qquad❖ \: \sf \:3.14 \times \frac{1}{4} [/tex]

[tex]\qquad❖ \: \sf \:0.785 \: \: m {}^{2} [/tex]

Area left for flowers :

[tex]\qquad❖ \: \sf \:6 - 0.785[/tex]

[tex]\qquad❖ \: \sf \:5.215 \: \: m {}^{2} [/tex]

[tex]\qquad❖ \: \sf \: \approx5.22 \: \: m {}^{2} [/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

The amount of the lease receivable will be $146999.

It should be noted that lease receivables means all the payments under the operative documents in respect of:

(i) the Purchaser Basic Rent,

(ii) the Maximum recourse amount,

(iii) the Purchaser Balance and

(iv) any purchase by the Lessee of the Properly up to the amount of the Capital then outstanding.

In this case, the lease is a 7 year period and requires equal annual payments of $26143 at the beginning of each year and the first payment is received on January 1, 2017.

The lease receivable will be:

= PVAD8%,7 × Lease payment

= 5.62288 × 26143

= $146999

In conclusion, the lease receivable is $146999.

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The solution to the question is 265.

Logarithm of a number A is the exponent or power or index n, a given number called the base B would be raised to give the number A.

So n is the logarithm of A to base B.

Analysis:

log 2( log2(x-9)) = 3

log2(x-9) = [tex]2^{3}[/tex]

log2(x-9) = 8

x-9 = [tex]2^{8}[/tex]

x-9 = 256

x = 256 + 9

x = 265

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Distribute the summation over the sum.

[tex]\displaystyle \sum_{i=1}^{101} (-5a_i - 12b_i) = -5 \sum_{i=1}^{101} a_i - 12 \sum_{i=1}^{101} b_i[/tex]

Now plug in the known sums and simplify.

[tex]\displaystyle \sum_{i=1}^{101} (-5a_i - 12b_i) = -5(-12) - 12(-19) = \boxed{288}[/tex]

Answer:

[tex]6 + ( \frac{4}{3} + ( \frac{5}{6} - \frac{1}{3} ))[/tex]

[tex] = 6 + ( \frac{4}{3} + ( \frac{5 - 2}{6})) [/tex]

[tex] = 6 + ( \frac{4}{3} + \frac{3}{6})[/tex]

[tex] = 6 + ( \frac{8 + 3}{6})[/tex]

[tex] = 6 + \frac{11}{6} [/tex]

[tex] = \frac{6}{1 } + \frac{11}{6} [/tex]

[tex] = \frac{36 + 11}{6} [/tex]

[tex] = \frac{47}{6} [/tex]

[tex] = 7 \frac{5}{6} [/tex]

Answer:

the area would be 176.7

Step-by-step explanation:

You know the area of a circle is A = pi×r^2, so you need the radius.

You know C = 2pi×r, so r = C/(2pi)

r = 15pi / 2pi = 7.5

A = pi(7.5)^2 = 56.25pi ~ 176.715 square inches.

The area of the circle is about 176.7 square inches.

The radius of the circle is 7.5 inches. Therefore, the area of the circle will be 56.25π square inches.

Let r be the radius of the circle. Then the area of the circle will be given as,

A = πr² square units

The circumference of the circle will be given as,

C = 2πr units

The circumference of a circle is 15π inches. Then the radius of the circle is calculated as,

15π = 2πr

r = 7.5 inches

The area of the circle is calculated as,

A = π x 7.5²

A = 56.25π square inches

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Answer:

C is 27 and d is 16

Step-by-step explanation:

count how many cubes are on the first level and multiple but how many it goes up too

Answer:

B

Step-by-step explanation:

When x = 1, it appears y is in between 1 and 2.

The only choice that satisfies this is B.

Answer:

The equation that represents the graph is y=2e^x

Step-by-step explanation:

plug in geogebra and confirm:D

Answer:

C 1/3

Step-by-step explanation:

out of the 6 possible moves you have the possibility of landing on question mark twice. 2/6 -> 1/3

The quotient of the given expression is 1/3

Fractions are written as a ratio of two integers

Given the expression below'

-1/9 / (-1/3)

Change the division to multiplication

-1/9 * -3/1

1/9 * 3/1

1/3

Hence the quotient of the given expression is 1/3

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Answer: A. ± [tex]\frac{1}{3} ,\frac{5}{3} ,1,3,5,15[/tex]

Step-by-step explanation:

± [tex]\frac{1, 3, 5, 15}{1, 3}[/tex]

= ± [tex]1, 3, 5, 15, 1/3, 5/3, 5[/tex]

Rearranging, we get:

= ± [tex]\frac{1}{3} ,\frac{5}{3} ,1,3,5,15[/tex]

The measure of angle C (m ∠C) is 40°

From the question, we are to determine the measure of angle C (m ∠C)

In any given triangle, the sum of all the angles is 180°

Thus,

In ΔABC, the angles sum up to 180°

That is,

∠A + ∠B + ∠C = 180°

From the given information,

m ∠A = 80°, m ∠B = 60°

Thus,

80° + 60° + m ∠C = 180°

140° + m ∠C = 180°

m ∠C = 180° - 140°

m ∠C = 40°

Hence, the measure of angle C (m ∠C) is 40°

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The identification of parts A,B andC is illustrated below with their various reasons given.

An equilateral triangle is the triangle that has all its sides equal in length and each angle is made up of angle 60°.

Part A = The similar triangle are RGE and PBE

Part B = The triangles selected are similar because it was formed by an interception of the parallel lines of the rectangle GCPR.

Part C= All the sides of the equilateral triangle are the same therefore the distance from B to E and from P to E is the same with BP which is 225ft.

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the value of sin ∅ is 12/ 13

From the given question,

We have,  cos ∅= 5/13

From the trigonometric identities, we have that

[tex]sin^2\alpha + cos^2\alpha = 1[/tex]

Then , let's substitute the value of cos ∅

[tex]sin^2\alpha +(\frac{-5}{13} )^2 = 1[/tex]

Let make sin the subject of formula and find the squares of the fraction

sin²∅ = [tex]1 - \frac{25}{169}[/tex]

Find the LCM

sin²∅ = [tex]\frac{169 - 25}{169}[/tex]

Find the difference

sin²∅ = [tex]\frac{144}{169}[/tex]

Find the square root

sin∅ = [tex]\sqrt{\frac{144}{159} }[/tex]

sin∅ = [tex]\frac{12}{13}[/tex]

In quadrant II , sin is positive, so we have

sin ∅ = 12/ 13

Thus, the value of sin ∅ is 12/ 13

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The number of unique passwords is 5400

The characteristics of the password are given as:

Since the characters in the password must unique, we have the following possibilities

So, the number of unique passwords is

Password = 5 * 10 * 4 * 9 * 3

Evaluate the product

Password = 5400

Hence, the number of unique passwords is 5400

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The average daily balance of the credit card is 2291.94

What is the average daily balance?

The average daily balance can be determined by dividing the total daily balance of month by numbers of days in month.

We can find average daily balance as shown below:

Average daily balance = total balance of the month/total number of days in the month

Total balance = (2500*10) +(2000*10) +(2750*5) +(2050*6)

= 25000+20000+13750+12300

= 71,050

Number of days = 31

Average daily balance = 71,050/31

= 2291.9354

Round to the nearest cent

= 2291.94

Hence, the average daily balance of the credit card is 2291.94

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