Identify whether the decimal expansion of each fraction is terminating or repeating

\(\huge\boxed{Hi!}\)

We are given 2 points that the line passes through, so we can determine its slope by using this formula:

\(\displaystyle\frac{y2-y1}{x2-x1}\)

However, there's a shortcut.

If you look at the two points, you will see that the x-coordinates of these 2 points are the same.

Let's plug in the numbers and see what we get:

\(\displaystyle\frac{9-5}{12-12}\)

Subtract:

\(\displaystyle\frac{4}{0}\)

Do you see what I see?

Remember, we cannot divide by zero, because division by 0 is

undefined.

Thus, the slope of the line is also undefined.

Now, is Phoebe right? Yes, sure!

The slope of the given line is undefined.

Here's something you should remember:

\(\bigstar\) If we are given 2 points, and their x-coordinates are the same, then the slope of the line is undefined.

\(\bigstar\) If we are given 2 points, and their y-coordinates are the same, then the slope of the line is 0.

Hope it helps! Enjoy your day!

~Just a determined gal

*The crown would be greatly appreciated ;)

\(\bf{erutaNtneliS\)

Answer:

0.51 or 51%.

Step-by-step explanation:

If there were 100 students in the school, then  38 have a dog 24 a cat and 11 have both.

So there are 38 - 11 = 27 who only have a dog and 24-11 = 13 who only have a cat.

So total number having a pet = 27+11+13

= 51 students.

So Prob(You will choose a student with a pet) = 51%.

Drawing a Venn diagram would help to answer this and make it  clearer.

Answer: answer above me is correct

Answer:

Step-by-step explanation:

5

c

Step-by-step explanation:

because i jist did the q

Answer:

61 c

Step-by-step explanation:

Circumference= 2πr = πd=3.14d

1*10 c+ 2*25 c+ 1*1 c = 61 c

Answer: 50x - 27

Step-by-step explanation:

To find the 8th term of the arithmetic sequence, we need to first find the common difference between consecutive terms:

Common difference (d) = second term - first term

d = (8x - 3) - (x + 1)

d = 7x - 4

Now, we can use the formula to find the nth term of an arithmetic sequence:

an = a1 + (n - 1)d

where a1 is the first term, d is the common difference, and n is the term number we want to find.

Plugging in the values, we get:

a8 = (x + 1) + (8 - 1)(7x - 4)

a8 = x + 1 + 7(7x - 4)

a8 = x + 1 + 49x - 28

a8 = 50x - 27

Therefore, the 8th term of the arithmetic sequence x + 1, 8x - 3, 15x - 7 is 50x - 27.

Answer:

X=13

Step-by-step explanation:

So if ΔABC ≅ ΔDEC, then that means that ∠B is ≅ ∠E. Therefore you would write the equation 3x=6x-39 and solve for X to get 13.

The surface area of the right rectangular prism 280 yd².

A rectangular prism is simply a three-dimensional solid shape which has six faces that are rectangles.

The surface area of a rectangular prism is expressed as;

S.A = 2( wl + hl + hw )

Where w is the width, h is height and l is length.

From the diagram:

Length l = 6 yd

Width w = 5 yd

Height = 10 yd

Plug the values into the above formula:

S.A = 2( wl + hl + hw )

S.A = 2( 5×6 + 10×6 + 10×5 )

S.A = 2( 30 + 60 + 50 )

S.A = 2( 90 + 50 )

S.A = 2( 140 )

S.A = 280 yd²

Therefore, the surface area is 280 square yards.

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ANSWER

EXPLANATION

We want to find the discrimininant of the equation:

The general form of a quadratic equaion is given as:

The discriminant is given as:

Therefore, we need to find that.

From the given equation:

Therefore, we have that the discriminant is:

That is the discriminant.

Since the discriminant is less than 0, then there are no real solutions of the equation.

Answer:

x = x, y = −5 −3x

Step-by-step explanation:

hope this helps

In a linear equation, 9.14 does he save by paying off the loan when the next payment is due .

What in mathematics is a linear equation?

An algebraic equation with simply a constant and a first-order (linear) term, such as y=mx+b, where m is the slope and b is the y-intercept, is known as a linear equation.

                               Sometimes, the aforementioned is referred to as a "linear equation of two variables," where x and y are the variables.

Jan  293  18  311  1,507 16.25%

Feb  296  15  311  1,212 32.67%

Mar 298           12  311   913 49.25%

Apr  301            9  311   612 66.00%

May  304    6  311  308 82.92%

Jun  308    3  311    0        100.00%

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The probability that fewer than 3 of them use their smartphones in meetings or classes is 0.0366.

The binomial probability distribution is a discrete probability law with two parameters: trial number (n) and success probability (p) (p). When there are two mutually exclusive outcomes of finite trials, the binomial distribution is utilised.

The probability that a specific 3 people all use their smartphones in meetings or class is 0.53³.

Probability that remaining 7 people do not use their phone is 0.47⁷

¹⁰C₃ = 10*9*8/(3*2*1) = 120.

Thus, the probability of exactly 3 people using their smartphones in meetings or in class is 10C3*0.533*0.477, or 0.0905.

So, adding up the 3 possibilities (that 0 of them, 1 of them, or 2 of them use their smartphones in meetings or class) we get

¹⁰C₀*0.530*0.4710 + ¹⁰C₁*0.531*0.479 + ¹⁰C₂*0.532*0.478 = 0.0366.

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Complete question:

Assume that when adults with smartphones are randomly​ selected, 53​% use them in meetings or classes. If 10 adult smartphone users are randomly​ selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes.

Answer:

E2020

Step-by-step explanation:

Answer:

d, supplementary, bc, addition

Step-by-step explanation:

Answer:

You will need 32 ounces

Step-by-step explanation:

2 pounds x 16 = 32 ounces

300 cal / 3 oz = x / 32 oz

(300 * 32) / 3 = 3,200 calories

The probability values  of the Poisson distribution are P(y ≤ 2) = 0.423 and P(y = 3) = 0.140

Probability 1

The given parameters here are:

P(y ≤ 2) when λ=3

The probability of a Poisson distribution is represented as

P(x) = λˣe⁻λ/x!

In this case,

x = y ≤ 2

This means that

y = 0, 1 and 2

So, we have

P(y ≤ 2) = P(0) + P(1) + P(2)

This gives

\(P(y \le 2) = \frac{3^0 \times e^{-3}}{0!} + \frac{3^1 \times e^{-3}}{1!} +\frac{3^2 \times e^{-3}}{2!}\)

Evaluate

\(P(y \le 2) = \frac{1 \times e^{-3}}{1} + \frac{3 \times e^{-3}}{1} +\frac{9 \times e^{-3}}{2}\)

Solving further, we have

P(y ≤ 2) = 0.050 + 0.149 +0.224

P(y ≤ 2) = 0.423

Probability 2

The given parameters here are:

P(y = 3) when λ = 5

The probability of a Poisson distribution is represented as

P(x) = λˣe⁻λ/x!

In this case,

x = y = 3

This gives

P(y = 3) = 5³ * e⁻⁵/3!

Evaluate

P(y = 3) = 125 * e⁻⁵/6

Solving further, we have

P(y = 3) = 0.140

Hence, the probability is 0.140

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Mr. Lowe baked a total of 17 cookies.

To find out, we can work backward from the one remaining cookie. Before Anna ate one cookie, there were 2 cookies left. Before Noah ate one half of the remaining cookies, there were 4 cookies left. Before Jacob and Emma each ate two thirds of what was left, there were 6 cookies left. Before Sarah ate half as many as Nathan, there were 9 cookies left. Before Nathan ate a half dozen (or 6) cookies, there were 15 cookies left. And before Mr. Lowe ate one cookie, there were 16 cookies left.
So, the total number of cookies Mr. Lowe baked is:
1 (remaining cookie) + 1 (Anna's cookie) + 2 (Noah's cookies) + 2 (Jacob's cookies) + 2 (Emma's cookies) + 3 (Sarah's cookies) + 6 (Nathan's cookies) + 1 (Mr. Lowe's cookie) = 17 cookies.
Therefore, Mr. Lowe baked a total of 17 cookies.

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Values of a and b do the function f(x) = ax/(b + x²) have a critical point at x = 2 and x = -2 are A = 1, b = 4. The correct answer is option e. We need to find the first derivative of the function and set it equal to zero at x = 2 and x = -2.

The primary subordinate of f(x) is:

f'(x) = a(b - x²)/((b + x²)²)

Setting f'(2) = 0, we get:  a(b - 4)/((b + 4)²) =

Since a cannot be zero, we must have: b = 4

Setting f'(-2) = 0, we get: a(b - 4)/((b + 4)²) =

Since a cannot be zero, we must have: b = 4

Hence, as it were conceivable esteem for a and b could be a = 1 and b = 4.

We are able to check that this choice of a and b works by computing the moment subsidiary of f(x) and confirming that it is negative at x = 2 and x = -2, which would affirm that we have found a local maximum and a neighborhood least, separately. The moment subordinate of f(x) is:

f''(x) = 2ax(b - 3x²)/((b + x²)³)

f''(2) = -16/27 <

f''(-2) = -16/27 <

Subsequently, the proper reply is e. A = 1, b = 4.

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The value of the integral is approximately 0.3144.

The Maclaurin series for sin(x) is:

sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...

So, for sin(2x^2), we can substitute x^2 with 2x^2 to get:

sin(2x^2) = 2x^2 - (2x^2)^3/3! + (2x^2)^5/5! - (2x^2)^7/7! + ...

Now, we can integrate this series term by term to get:

∫₀^(0.8) sin(2x^2) dx = [2x^3/3 - (2x^7)/3(3!) + (2x^11)/5(5!) - (2x^15)/7(7!) + ...]₀^(0.8)

Evaluating this at x=0.8 and x=0, we get:

∫₀^(0.8) sin(2x^2) dx = [2(0.8)^3/3 - (2(0.8)^7)/3(3!) + (2(0.8)^11)/5(5!) - (2(0.8)^15)/7(7!)] - [0]

Simplifying this, we get:

∫₀^(0.8) sin(2x^2) dx ≈ 0.3144

Therefore, the value of the integral is approximately 0.3144.

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Data can be divided on the basis of qualitative nature into two types as follows:

Thus, it is clear that nominal data can not be ranked or ordered but ordinal data are the type of qualitative data that are categorised and then ranked or ordered.

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The line determined by points P and vector v is unique, but P and v themselves are not uniquely determined by the line.

Given a line l determined by a point P and a vector v, it is possible to have different combinations of P and v that yield the same line.

To understand this, let's consider a simple example in a two-dimensional plane. Suppose we have two points P1(1, 1) and P2(2, 2) and their corresponding vectors v1(1, 0) and v2(2, 0). Both sets of points and vectors lie on the same line y = x, as the vectors v1 and v2 have the same direction. Thus, we have two different combinations of P and v that determine the same line.

In a more general setting, the direction of the vector v determines the orientation of the line, while the point P determines the position of the line in space. If we keep the direction of v constant and change the position of P, we obtain different lines that are parallel to each other. Similarly, if we keep the position of P constant and change the direction of v, we obtain lines with different orientations that pass through the same point.

Therefore, while the line determined by points P and vector v is unique, P and v themselves are not uniquely determined by the line. Different combinations of P and v can yield the same line, leading to multiple possibilities for the specific values of P and v given a line.

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Activity 1.a - Identifying Differences between Cash and Accrual Basis Cash basis accounting and accrual basis accounting are two methods of accounting used in bookkeeping to keep track of the income and expenses of a company or organization.

The following table lists the differences between cash basis accounting and accrual basis accounting based on Johnny Flowers Law Firm's advertising prepayment scenario. Cash Basis Accrual Basis Cash Payment January 1, $510Advertising expenses recorded on January 1,

$510Expenses Recorded January V $0Expenses Recorded January V $0January 31 $0Expenses Recorded January V $0February 28 $0Expenses Recorded January V $0March 31 $0Expenses Recorded January V $0April 30 $0Expenses Recorded January V $0May 21 $0Expenses Recorded January V $0June 3 $0Expenses Recorded January V $0Total Expenses Recorded $510.

Total Expenses Recorded $510Cash basis accounting records revenue and expenses only when they are received or paid, while accrual basis accounting records revenue and expenses when they are incurred. In the case of Johnny Flowers Law Firm's advertising prepayment scenario, cash basis accounting would show $510 in expenses recorded in January when the payment was made, and $0 in expenses recorded in the following months, while accrual basis accounting would show $510 in expenses recorded in January, February, March, April, May, and June because that is when the advertising is incurred or used.

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Answer:

x = 15

Step-by-step explanation:

As given, the equation 40 = 25 + x

As on the R.H.S of the equation, there is one constant and a variable x

As to isolate the variable we have to subtract the same constant so that that constant gives 0.

As we have the constant value 25

So, subtract 25 from both side of the equation, we get

40 - 25 = 25 + x - 25

⇒15 = x

∴ we get

x = 15

Answer:

12) x = 49.5

13) x = 8.5

14) x = 44.8

15) x = 4.6

Step-by-step explanation:

We can solve for the missing side lengths using trigonometry

Remember these are the trigonometric ratios

Sin = Opposite over Hypotenuse

Cos = Adjacent over Hypotenuse

Tan = Opposite over Adjacent

For #12

We can use trig ratio tangent as we are given an angle with a measure of 68° and its adjacent side length and we need to find the opposite side length. When dealing with opposite and adjacent we can use tan.

That being said we want to create an equation to solve for x

\(tan(68)=\frac{x}{20}\)

Now we solve

* multiply each side by 20 *

\(20tan(68) = 49.5\\\frac{x}{20} 20=x\\\)

we're left with x = 49.5

It seems that all of the problems are dealing with an angle, its opposite side and its adjacent side. So we will use the same trig ratio "tan" and process to solve for x.

For 13)

\(tan(62)=\frac{16}{x} \\\)

*multiply each side by x*

\(xtan(62)=16\)

*divide each side by tan(62)

\(\frac{xtan(62)}{tan(62)} =x\\\frac{16}{tan(62)} =8.5\)

we're left with x = 6.5

For 14)

\(tan(15)=\frac{12}{x}\)

multiply each side by x

\(xtan(15)=12\)

divide each side by tan15

\(\frac{xtan(15)}{tan(15)} =x\\\frac{12}{tan(15)} =44.8\)

we're left with x = 44.8

For 15)

\(tan(74)=\frac{16}{x}\)

multiply each side by x

\(xtan(74)=16\)

divide each side by tan(74)

\(\frac{xtan(74)}{tan74} =x\\\frac{16}{tan(74)} =4.6\)

we're left with x = 4.6

The value of the discriminant for the equation "2x² + 5x + 3"  is 1 , and the it has 2 real and rational roots .

The Discriminant of a quadratic equation of the form of ⇒ "ax² + bx + c" is calculated  by the expression "b² - 4ac"  .

the values of the variables "a", "b" and "c" ; for quadratic equation "2x² + 5x + 3" ,

the value of  a is = 2, b is = 5, and c is = 3.

substituting values ;

So , the discriminant is ⇒ b² - 4ac

⇒ 5² - 4×2×3  ;

⇒ 25 - 24

⇒ 1

Since , the discriminant is (1)positive, which means that the equation has 2 real roots. and since, discriminant is a perfect square, that means the roots are rational numbers.

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The given question is incomplete , the complete question is

Find the value of the discriminant for the equation 2x² + 5x + 3 , and then describe the number and type of roots for the equation .

Answer:

4.

Step-by-step explanation:

We have proven that no two events a and b can be independent unless either a or b is the whole sample space or the empty set below.

A sample space can be defined as the set of all possible outcomes of any experiment.

We are already aware that the outcomes of this die are going to be (1,2,3,..p). If we have a pair of proper events A, B which are independent, this would mean that = A∩B={∅}  (i)

This is a contradiction. So, now we suppose that = A∩B=C, for some proper event C (ii)

Using the values of (i) and (ii), we get -

= |C|p=P(A∩B)=P(A)P(B)=|A||B|p2.

= p|C|=|A||B|.

From this, we understand that neither A nor B are going to be full spaces and hence, no two events a and b are independent and 0<|A|<p and 0<|B|<p. Since |C| is also going to be less than 0 and p is prime, p divides either |A| or |B|.  if p was not prime (say, p=4), then you would only need the factors of p to divide either |A|,|B|, so, it becomes necessary for p to be a prime number. Hence, proved.

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Answer:

it is true

Step-by-step explanation:

The Area of annulus - the region between the two circle is 50.26cm²

According to the question,

The length of chord of the large circle which is tangent to the small circle : AB = 8cm

Le the radius of large circle "R" and radius of smaller circle "r"

AB is a tangent to the smaller circle

AM = 8/2

=> AM = 4cm

In ΔOAM,

Using Pythagoras theorem ,

OA² = OM² + AM²

=> R² = r² + (4)²

=> R² - r² = 16

So, the area of annulus = π(R² - r²)

=> 16π (putting π = 3.14)

=> 50.26cm² is required area

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Answer:

800

Hope this helps!

a)  The Taylor polynomial of degree 2 for f(x) = sec(x) centered at a = 0 is:

T2(x) = 1 + 0(x-0) + (1/2)(2)(x-0)^2

T2(x) = 1 + x^2

b)   The interval is [-0.1,0.1], we can take the maximum value of |x| to be 0.1. Thus,

|R2(x)| ≤ 0.25229 (rounded to six decimal places).

(a) The Taylor polynomial of degree 2 for f(x) = sec(x) centered at a = 0 is given by:

T2(x) = f(a) + f'(a)(x-a) + (1/2)f''(a)(x-a)^2

Since a=0 and f(x) = sec(x), we have:

f(0) = sec(0) = 1

f'(x) = sec(x)tan(x)

f'(0) = sec(0)tan(0) = 0

f''(x) = sec(x)tan^2(x) + sec(x)

f''(0) = sec(0)tan^2(0) + sec(0) = 2

Therefore, the Taylor polynomial of degree 2 for f(x) = sec(x) centered at a = 0 is:

T2(x) = 1 + 0(x-0) + (1/2)(2)(x-0)^2

T2(x) = 1 + x^2

(b) Taylor's Inequality states that if |f^(n+1)(c)| ≤ M for all x in the interval [a,x] and some constant M, then the remainder term Rn(x) satisfies the inequality:

|Rn(x)| ≤ M/[(n+1)!]|x-a|^(n+1)

In this case, we need to estimate the maximum value of the third derivative of f(x) = sec(x) on the interval [-0.1,0.1]. We have:

f'''(x) = sec(x)[3tan^2(x)+sec^2(x)]

Since sec(x) is always positive and increasing on the interval, we only need to consider the maximum value of 3tan^2(x)+sec^2(x) on the interval. This occurs at x = 0.1, and we have:

3tan^2(0.1)+sec^2(0.1) ≈ 9.025

So, we can take M = 9.025.

Using n = 2 and a = 0 in Taylor's Inequality, we get:

|R2(x)| ≤ 9.025/[(2+1)!]|x-0|^(2+1)

|R2(x)| ≤ 9.025/6|x|^3

Since the interval is [-0.1,0.1], we can take the maximum value of |x| to be 0.1. Thus,

|R2(x)| ≤ 0.25229 (rounded to six decimal places).

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Answer:

x = \(\sqrt{3}\)

Step-by-step explanation:

the ratio of side to hypotenuse in a right triangle is 1 : \(\sqrt{2}\)

in this problem we can use 'x' in place of 1 and \(\sqrt{6}\) in place of \(\sqrt{2}\) to create a proportion like this:

1/x = \(\sqrt{2}\)/\(\sqrt{6}\)

cross-multiply to get:

\(\sqrt{2}\)·x = \(\sqrt{6}\)

x = \(\frac{\sqrt{6}}{\sqrt{2}}\)

x = \(\sqrt{3}\)