Give an example of each of the following, or argue that such a request is impossible.

a. A sequence that has a subsequence that is bounded but contains no subsequence that converges.
b. A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values.
c. A sequence that contains subsequences converging to every point in the infinite set {1, 1/2, 1/3, 1/4, 1/5,...}.
d. A sequence that contains subsequences converging to every point in the infinite set {1, 1/2, 1/3, 1/4, 1/5,...}, and no subsequences converging to points outside of this set.

Answer:

I guess how many times out of x they will do something.

for example how many times out of 10 will bob ask out his crush vs Joe

Step-by-step explanation:

There isn't really an easy way to measure this.

The scale of measurement most reasonably applies to Confidence in Education would be the Ordinal scale.

The scale of measurement deals with how the variables are defined and categorized.

There are 4 types of scales of measurement;

1. Nominal scale

2. Ordinal scale

3. Interval scale

4. Ratio scale

We know that the Ordinale scale is used for variables in ranked order but the difference between them could not be determined.

Hence, The scale of measurement most reasonably applies to Confidence in Education would be the Ordinal scale.

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Answer: t<40

8p+<184

Step-by-step explanation:

The inequalities representing the given condition will be [tex]t\leq40[/tex] and [tex]8p+t\leq184[/tex].

Given information:

A baker makes apple tarts and apple pies each day.

The baker receives a shipment of 184 apples every day.

A tart requires 1 apple and a pie requires 8 apples.

Let t be the number of tarts made each day and p be the number of pies made each day.

The baker makes no more than 40 tarts per day. So, the condition can be written as,

[tex]t\leq40[/tex]

And the condition for each day apple delivery can be written as,

[tex]8p+t\leq184[/tex]

Therefore, the inequalities representing the given condition will be [tex]t\leq40[/tex] and [tex]8p+t\leq184[/tex].

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9514 1404 393

Answer:

  19

Step-by-step explanation:

The prime factorization of 4221462 is ...

  4221462 = 2·3·7·100511

so, the 1-digit divisors are ...

  1, 2, 3, 6, 7

The total of these numbers is 19.

Answer:

10

Step-by-step explanation:

25 times .60 then take that answer and subtract it from 25 or half of 25 which it 12.5 then take off another 10%

Step-by-step explanation:

There are 12 months in a year so,

114/12= 9.5$/month

t=9.5m

Since T represents the total cost, we want to express it in terms of months by multiplying m which refers to the number of months by the monthly fee which i have already calculated.

Using this equation we can find out the answer to your previous question.

t=9.5m

but m=3×12+5=41 months (3 years 12 months each)

t=9.5(41)

t=389.5 $

Answer:

3 hours

Step-by-step explanation:

as jen took 3 hours maryana will take equal amount of time as jen

Well, we already know that it took Jen, 3 hours, by herself to complete half the work. Then, together they both completed it in half an hour. Now, you must summarize your given data.

3 hours = Half of the work for Jen

30 Minutes = Jen and Maryna Together To Finish

If the two girls were working at the same pace, it would take them an hour and a half. However, it only took them 30 minutes to complete that. Now, we realize that Maryna works 3x faster than Jen does.

The question is asking, how long would it take her to FINISH the job.

So that means, we must divide 3 by Jen's time.

3 / 3 = 1

So, that means that it would only take Maryna 1 hour to complete the rest of the job.

Answer:

Step-by-step explanation:

Hello,

First of all, let s take x real number different from 5, as we cannot divide by 0

[tex]\dfrac{x+5}{x-5}\geq 0 \\ \\<=> (x+5\geq 0 \ and \ x-5> 0) \ or \ (x+5\leq 0 \ and \ x-5< 0) \\ \\<=> (x\geq -5 \ and \ x> 5) \ or \ (x\leq -5 \ and \ x< 5)\\ \\<=> x> 5 \ or \ x\leq -5[/tex]

So the solution is

[tex]\Large \boxed{\sf \bf ]-\infty;-5]\cup]5;+\infty[}[/tex]

Thanks

Answer:

(−∞,−5]∪(5,∞)

Answer: Look at picture

Answer:

a.) 0,51

b.) 51%

c.) 1/100

d.) 1%

e.) 3/20

f.) 0,15

Step-by-step explanation:

[tex]a.) \: \frac{51}{100} = 51 \div 100 = 0.51[/tex]

[tex]b.) \: \frac{51}{100} = 51\%[/tex]

[tex]c.) \: 0.01 = \frac{1}{100} [/tex]

[tex]d. )\: 0.01 = \frac{1}{100} = 1\%[/tex]

[tex]e.) \: 15\% = \frac{15}{100} = \frac{3}{20} [/tex]

[tex]f.) \: 15\% = \frac{15}{100} = 0.15[/tex]

===============================

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Answer:

4x^3

Step-by-step explanation:

GCF is the Greatest Common Factor

12x^6 + 4x^4 + 28x^3

= 4x^3(3x^3 + x + 7)

GCF is 4x^3

Answer:

Let's solve your equation step-by-step.

p=(0.5)(2.718282)−5

Step 1: Simplify both sides of the equation.

p=(0.5)(2.718282)−5

p=1.359141+−5

p=(1.359141+−5)(Combine Like Terms)

p=−3.640859

p=−3.640859

Answer:

p=−3.640859

Step-by-step explanation:

Gradient = Slope.

Use two points on the line to find the slope.

Slope = change in y over change in x

Points: (-4,0) and (0,-4)

Slope = (-4 - 0) / (0 - -4)

Slope = -4/4 = -1

Slope = -1

Answer:

Question (a)

[tex]x^2-a^2-bx-ab=0[/tex]

Write in standard form [tex]ax^2+bx+c=0[/tex]:

[tex]\implies x^2-bx-(a^2+ab)=0[/tex]

Therefore,

Using quadratic formula:

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\implies x=\dfrac{b\pm\sqrt{(-b)^2+4(1)(a^2+ab)} }{2(1)}[/tex]

[tex]\implies x=\dfrac{b\pm\sqrt{b^2+4a^2+4ab} }{2}[/tex]

[tex]\implies x=\dfrac{b\pm\sqrt{(b+2a)^2} }{2}[/tex]

[tex]\implies x=\dfrac{b \pm (b+2a)} {2}[/tex]

[tex]\implies x=\dfrac{2b+2a} {2}=b+a[/tex]

[tex]\implies x=\dfrac{-2a} {2}=-a[/tex]

Question (b)

[tex]6x^2-15ax=2bx-5ab[/tex]

Write in standard form  [tex]ax^2+bx+c=0[/tex]:

[tex]\implies 6x^2-15ax-2bx+5ab=0[/tex]

[tex]\implies 6x^2-(15a+2b)x+5ab=0[/tex]

Therefore,

Using quadratic formula:

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{(-15a-2b)^2-4(6)(5ab)} }{2(6)}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{225a^2+60ab+4b^2-120ab} }{12}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{225a^2-60ab+4b^2} }{12}[/tex]

[tex]\implies x=\dfrac{(15a+2b)\pm\sqrt{(15a-2b)^2} }{12}[/tex]

[tex]\implies x=\dfrac{(15a+2b) \pm (15a-2b) }{12}[/tex]

[tex]\implies x=\dfrac{30a }{12}=\dfrac52a[/tex]

[tex]\implies x=\dfrac{4b}{12}=\dfrac13b[/tex]

Question (c)

[tex]\dfrac{x^2}{x-1}=\dfrac{a^2}{2a-4}[/tex]

Cross multiply:

[tex]x^2(2a-4)=a^2(x-1)[/tex]

Write in standard form  [tex]ax^2+bx+c=0[/tex]:

[tex](2a-4)x^2-a^2x+a^2=0[/tex]

Therefore,

Using quadratic formula:

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\implies x=\dfrac{a^2\pm\sqrt{(-a^2)^2-4(2a-4)(a^2)} }{2(2a-4)}[/tex]

[tex]\implies x=\dfrac{a^2\pm\sqrt{a^4-8a^3+16a^2} }{4a-8}[/tex]

[tex]\implies x=\dfrac{a^2\pm\sqrt{a^2(a-4)^2} }{4a-8}[/tex]

[tex]\implies x=\dfrac{a^2\pm a(a-4)}{4a-8}[/tex]

[tex]\implies x=\dfrac{a^2\pm (a^2-4a)}{4a-8}[/tex]

[tex]\implies x=\dfrac{2a^2-4a}{4a-8}=\dfrac{2a(a-2)}{4(a-2)}=\dfrac12a \ \ \ (a\neq 2)[/tex]

[tex]\implies x=\dfrac{4a}{4a-8}=\dfrac{a}{a-2} \ \ \ \ (a\neq 2)[/tex]

The door of the ramp must begin at a distance of 25.72ft in order to raise the ramp by 5ft.

Data;

To solve this problem, this is a right angle triangle and we can use trigonometric ratio SOHCAHTOA here.

Since we have the value of angle and opposite and solving only for adjacent, we can use tangent of the angle.

[tex]tan \theta = \frac{opposite}{adjacent} \\[/tex]

Let's substitute the values into the formula above and solve

[tex]tan 11 = \frac{5}{x} \\x = \frac{5}{tan 11} \\x = 25.72ft[/tex]

The door of the ramp must begin at a distance of 25.72ft in order to raise the ramp by 5ft.

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Answer:

Step-by-step explanation:

Opposite sides of a parallelogram are congruent, so:

[tex]8x+3=19 \longrightarrow x=2[/tex]

[tex]27=-y-5 \longrightarrow y=-32[/tex]

Use the old exp-log trick and properties of the logarithm to rewrite the limit as

[tex]\displaystyle \lim_{n\to\infty} \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)^{\frac xn}[/tex]

[tex]\displaystyle = \exp\left[\lim_{n\to\infty} \frac xn \ln \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac nk\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\left(\frac nk\right)^2\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac1{k/n}\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\frac1{\left(k/n\right)^2}\right)\right][/tex]

The first limit converges to 0, since n! asymptotically behaves like nⁿ, so ln(nⁿ/n!) → ln(1) = 0.

The two remaining sums converge to definite integrals:

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x + \frac1{\frac kn}\right) = \int_0^1 \ln\left(x + \frac1y\right) \, dy = \frac{(x+1) \ln(x+1)}x[/tex]

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x^2 + \left(\frac1{\frac kn}\right)^2\right) = \int_0^1 \ln\left(x^2 + \frac1{y^2}\right) \, dy = \frac{2\tan^{-1}(\sqrt x)}{\sqrt x} + \ln(1+x)[/tex]

It follows that

[tex]f(x) = \exp\left[x\left(0 + \dfrac{(x+1)\ln(x+1)}x - \dfrac{2\tan^{-1}(\sqrt x)}{\sqrt x} - \ln(x+1)\right)\right][/tex]

[tex]f(x) = \exp\left[\ln(x+1) - 2\sqrt x \tan^{-1}(\sqrt x)\right][/tex]

[tex]f(x) = (x+1) e^{-2\sqrt x \tan^{-1}(\sqrt x)}[/tex]

By computing f'(x) and f''(x), it's easy to show that f'(x) ≤ 0 and f''(x) ≥ 0 for all x > 0. So f(x) is decreasing and f'(x) is increasing, and

• (A) f(1/2) ≥ f(1) is true

• (B) f(1/3) ≤ f(2/3) is false

• (C) f'(2) ≤ 0 is true

• (D) f'(3)/f(3) ≥ f'(2)/f(2)   ⟺   f'(3)/f'(2) ≥ f(3)/f(2)   ⟺   (something larger than 1) ≥ (something smaller than 1) is true

Answer:

See below

Step-by-step explanation:

Definitions are:

Answer:

fog =gofthen it will be equal

Answer:

Angles 7 and 8

Step-by-step explanation:

They are next to each other, and share a common side.

Answer:

5

Step-by-step explanation:

6 / 2 = 15 / l

3 = 15 / l

l = 15 / 3

l = 5

Answer:

6

Step-by-step explanation:

i took the test

The height of the prism is 49.2 feet

The given parameters are:

The area of the trapezoid is:

Area = 0.5 * (Sum of opposite sides) * Height

So, we have:

1280 = 0.5 * 52 * Height

This gives

1280 = 26 * Height

Divide both sides by 26

Height = 49.2

Hence, the height of the prism is 49.2 feet

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Answer:

[tex]x=20[/tex]

Step-by-step explanation:

[tex]x^2=8(8+42)[/tex]

[tex]x^2=8*50[/tex]

[tex]x=20[/tex]

~

Answer: 5/9 is the simplified fraction for 25/45 by using the GCD or HCF method. Thus, 5/9 is the simplified fraction for 25/45 by using the prime factorization method.

Answer:

32 Bags.

Step-by-step explanation:

Answer:

B. 75.5 square inches

Step-by-step explanation:

Area (pentagon) = Area (rectangle) - Area (triangle)

= length x width - 1/2 x breadth x height

= 10 x 8 - 1/2 x 3 x 3

= 80 - 4.5

= 75.5 square inches

Solution :-

B. 75.5 square inches

Answer:

B) 75.5 in²

Step-by-step explanation:

Area of rectangle = width × length

⇒ area of rectangle = 8 × 10 = 80 in²

Area of triangle = 1/2 × base × height

⇒ area of triangle = 1/2 × 3 × 3 = 4.5 in²

Area of pentagon = area of rectangle - area of triangle

                              = 80 - 4.5

                             = 75.5 in²

Answer: B 0.4

Step-by-step explanation:

50-20=30

0.4 is 40%

40% of 50 is 20

Answer:

x=3

Step-by-step explanation:

In(x-1) = In(6) - In(x)

In(x-1) + In(x) = In(6)

In((x-1)x) = In(6)

In(x^2-x) = In(6)

x^2-x=6

x^2-x-6=0

x^2+2x-3x-6=0

x(x+2)-3(x+2)=0

(x-3)(x+2)=0

x=3 because x=-2 need to be cancel out