write any five applications of cyliner​

Both arguments are valid.

The validity of the arguments can be determined by using the concept of propositional logic.

This argument can be represented as: (P∨D∨B)∧(D∨C∨B). The form of the argument is p ∧ q, which is a valid form.

This argument can be represented as: P∨C. The form of the argument is p ∨ q, which is also a valid form.

So, both arguments are valid. The validity of an argument depends on the form of the argument and not the specific proposition used. In both cases, the form of the argument is valid, so the argument is also valid.

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Answer:

The slope of g(x) is less than the slope of f(x)

Step-by-step explanation:

Find the slopes of both lines with rise/run

f(x): 3/1 = 3

g(x): 2/4 = 1/2

The slope of g(x) is less than the slope of f(x)

Answer:

B. The slope of g(x) is less than the slope of f(x)

Step-by-step explanation:

The slope of f(x) is 2/1 or just 2 and the slope of g(x) is 1/2, less than f(x).

The answers that describe the quadrilateral DEFG area rectangle and parallelogram.

The correct answer choice is option A and B.

A quadrilateral is a parallelogram, which has opposite sides that are congruent and parallel.

Quadrilateral DEFG

if line DE || FG,

line EF // GD,

DF = EG and

diagonals DF and EG are perpendicular,

then, the quadrilateral is a parallelogram

Hence, the quadrilateral DEFG is a rectangle and parallelogram.

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The shortest distance from point A to the circle is zero. In other words, point A lies on the circle itself.

Length of the tangent, LL = (4/3) * r, where r is the radius of the circle.

We need to find the shortest distance from point A to the circle.

The shortest distance from an external point to a circle is along the line connecting the point and the center of the circle, perpendicular to the tangent.

Let's assume that the center of the circle is point O.

Now, consider the right triangle formed by the line segment AO, the radius of the circle (r), and the shortest distance (h) from point A to the circle.

By Pythagoras' theorem, we have:

AO^2 = h^2 + r^2

Since AO is the radius of the circle, it is equal to r:

r^2 = h^2 + r^2

Subtracting r^2 from both sides, we get:

0 = h^2

This implies that h = 0.

As a result, zero is the shortest distance between point A and the circle. Point A is therefore located on the circle itself.

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The algebraic equation of the verbal statement is (x + 2) / 5 = 8 and the value of x is 38.

Algebraic Equation

An algebraic equation means the statement of the equality of two expressions formulated by applying to a set of variables the algebraic operations, namely, addition, subtraction, multiplication, division, raising to a power, and extraction of a root.

Given,

The quotient of two more than a number and five is eight.

Here we  need to translate the given verbal statement into an algebraic expression and solve it.

The quotient of two more than a number and five is eight.

Let, the number be x.

According to the question, the equation is

=> (x + 2) / 5 =8

=> x + 2 = 8 x 5

=> x + 2 = 40

=> x = 40 - 2

=> x = 38

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P = (3,-4)

M = (1,3)

Q = (Qx,Qy)

Midpoint = (x1+x2)/2 , (y1+y2)/2

Mx = (Px+Qx) /2

1= (3+Qx) /2

Solve for Qx

1 (2) = 3+Qx

2 = 3 +Qx

2-3 = Qx

Qx= -1

My = (y1+y2) /2

3 = (-4+Qy) /2

Solve for QY

3 (2) = -4+Qy

6= -4 + Qy

6+4 = Qy

10 = Qy

Coordinates of the Endpoint Q:

Q = (-1,10)

Answer:

\($1,800\)

Answer:

g= 92°

h= 88°

k= 91°

m= 89°

Step-by-step explanation:

Answer:

g = 92°

h = 88°

k = 91°

m = 89°

Step-by-step explanation:

Have a nice day!

The two sides of an inequality are not the same so inequality can never be true.

When two expressions are connected by a sign like "not equal to," "greater than," or "less than," it is said to be inequitable. The inequality shows the greater than and less than relationship between variables and the numbers.

Given that the inequality is 2(3x - 1) +4> 6x + 2. The inequality will be solved as below;-

2(3x - 1) +4> 6x + 2

6x - 2 + 4 > 6x + 2

6x + 2 > 6x + 2

The two sides are equal so inequality can never be true.

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The option that is not a benefit of stretching is 1. Poor circulation

The option that is not in upper body stretch is 3. Straddle stretch

The benefits include:

The straddle stretch is one that is not considered an upper body stretch. Straddle stretch, also known as the middle split, requires a lot of leg flexibility. In conclusion, the correct option is C.

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The number of small tiles are needed​ is 300 tiles.

Using area of rectangular formula

Area of the room = length(l) × breadth (b)

Area of the room=300 cm×180 cm

Area of the room=54,000 cm²

Number of tiles needed = Area of rectangular region / Area of one tile

Number of tiles needed=54,000/180

Number of tiles needed=300 tiles

Therefore the number of small tiles are needed​ is 300 tiles.

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Answer:

10:12, 5:24, 9:13+1/3, 3:40, 20:6, 7:17+1/7

Step-by-step explanation:

You simply divide 120 by each of the numbers.

120/10=12, 120/5=24, 120/9=40/3, 120/3=40, 120/20=6, 120/7=12+1/7

Answer:6

Step-by-step explanation:

Let the number be x.

Now, 3x+6=24

3x=24-6

3x=18

x=18/3

x=6

Answer:

6

Step-by-step explanation:

We can work the problem backwards

Since something gives 24 that means it equals 24

= 24

Something increased by 6, so something added by 6

Something + 6 = 24

Making something = x

x + 6 = 24

If we subtract 6 from both sides

x = 18

Trice a number, so divide this number by 3

x = 6

Trice of 6 is 18... 18 increased by 6 gives 24

It should be noted that C = π(R + 1)² × 20 is an expression for the estimated cost of the base.

The surface area of the base is given by

A = πr²

where r is the radius of the base. Since the radius of the base is 1 foot larger than the radius of the cylinder, we have

r = R + 1

Substituting this into the expression for the area of the base gives

A = π(R + 1)²

The cost of the base is given by

C = A * 20

C = π(R + 1)² * 20

This is an expression for the estimated cost of the base.

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Answer:

No Solution.

Step-by-step explanation:

For starters, here are our two equations:

\((x^2-x-6)>0\)

\(10-2x<5\)

The first option is to solve for x in our second equation. First, we add 2x to both sides to get \(10<5+2x\). Then, we subtract 5 from both sides to get \(2x<5\). Finally, we divide both sides by 2 to get \(x<2.5\). We can plug in 2.5 into our second equation to test if it works. If it does not work, then anything less than 2.5 won't work, and the solution would be impossible.

Now we can plug in our values into our first equation. We now have the equation \(2.5^2-2.5-6>0\). We can simplify to get \(-2.25>0\). This test failed, so there are no solutions. A solution would be impossible.

Given formula: \(\sf A_n= -4(2)^{n-1}\)

simplify substitute the number of term by replacing n

Find first five terms:

\(\sf A_1= -4(2)^{1-1}\)  ⇒  \(\sf A_1= -4\)

\(\sf A_2= -4(2)^{2-1}\)  ⇒  \(\sf A_2= -8\)

\(\sf A_3= -4(2)^{3-1}\)  ⇒  \(\sf A_3= -16\)

\(\sf A_4= -4(2)^{4-1}\)⇒  \(\sf A_4= -32\)

\(\sf A_5= -4(2)^{5-1}\)⇒  \(\sf A_5= -64\)

Tenth term:

\(\sf A_{10}= -4(2)^{10-1}\)   ⇒  \(\sf A_{10}= -2048\)

Answer:

\(\sf a_1=-4\\a_2=-12\\a_3=-20\\a_4=-28\\a_5=-36\\a_{10}=-76\)

Step-by-step explanation:

\(\textsf{Given sequence}: \sf a_n=-4(2n-1)\)

This formula is for the nth term of the sequence.

Therefore, to find any term of the sequence, substitute the position of the term you wish to find as n.

For example, to find the 10th term, substitute n = 10 into the formula:

\(\begin{aligned}\sf a_{10} & =-4[2(10)-1)]\\ & = -4(20-1)\\ & = -4(19)\\ & = -76\end{aligned}\)

To find the first 5 terms, substitute n = 1 through n = 5 into the formula:

\(\sf a_1=-4[2(1)-1]=-4(2-1)=-4(1)=-4\)

\(\sf a_2=-4[2(2)-1]=-4(4-1)=-4(3)=-12\)

\(\sf a_3=-4[2(3)-1]=-4(6-1)=-4(5)=-20\)

\(\sf a_4=-4[2(4)-1]=-4(8-1)=-4(7)=-28\)

\(\sf a_5=-4[2(5)-1]=-4(10-1)=-4(9)=-36\)

Answer:

G

Step-by-step explanation:

a + b + c = 80

Express a and c in terms of b using the ratios in fractional form, that is

\(\frac{a}{b}\) = \(\frac{2}{3}\) ( cross- multiply )

3a = 2b ( divide both sides by 3 )

a = \(\frac{2}{3}\) b

\(\frac{b}{c}\) = \(\frac{3}{11}\) ( cross- multiply )

3c = 11b ( divide both sides by 3 )

c = \(\frac{11}{3}\) b

Substitute the values for a and c into the original equation

\(\frac{2}{3}\) b + b + \(\frac{11}{3}\) b = 80 ( multiply through by 3 to clear the fractions )

2b + 3b + 11b = 240 , that is

16b = 240 ( divide both sides by 16 )

b = 15 → G

Answer:

I think it's C.

Step-by-step explanation:

Answer:

D. 17

Step-by-step explanation:

No-

Answer:

Step-by-step explanation:

4x^2-40-120

4x^2-160

4x^2=160

x^2=40

sqrt40=6.32

P cannot be bounded below by any multiple of ||·||₂, since ||f||₂ can be

Let's first examine whether p(f) satisfies the definition of a norm.

To be a norm, p(f) must satisfy the following properties for all f ∈ L²[0, 1]:

Non-negativity: p(f) ≥ 0

Definiteness: p(f) = 0 if and only if f = 0 almost everywhere

Homogeneity: p(αf) = |α|p(f) for any scalar α

Triangle inequality: p(f+g) ≤ p(f) + p(g) for any f, g ∈ L²[0, 1]

Non-negativity:

We have - (11/1²) ≤ 0 for all f, and clearly ||f||₂ ≥ 0 as well. Therefore, p(f) is non-negative.

Definiteness:

If p(f) = 0, then we must have f = 0 almost everywhere, since the constant function 1 belongs to L²[0, 1] and has a nonzero integral. Conversely, if f = 0 almost everywhere, then p(f) = 0. Therefore, p(f) satisfies the definiteness property.

Homogeneity:

For any scalar α, we have:

p(αf) = - (11/1²) + ||αf||₂¹ + ∫₀¹ |αf'|² dx

= - (11/1²) + |α|²||f||₂¹ + ∫₀¹ |αf'|² dx

= |α|(- (11/1²) + ||f||₂¹ + ∫₀¹ |f'|² dx)

= |α|p(f),

so p(f) satisfies the homogeneity property.

Triangle inequality:

For any f, g ∈ L²[0, 1], we have:

p(f+g) = - (11/1²) + ||f+g||₂¹ + ∫₀¹ |(f+g)'|² dx

= - (11/1²) + ||f||₂¹ + ||g||₂¹ + 2∫₀¹ |f'g'| dx + ∫₀¹ |g'|² dx + ∫₀¹ |f'|² dx

≤ - (11/1²) + ||f||₂¹ + ||g||₂¹ + 2(||f'||₂ ||g||₂ + ||f||₂ ||g'||₂) + ||f'||₂² + ||g'||₂² + ||f||₂² + ||g||₂²

≤ - (11/1²) + 2(||f||₂¹ + ||g||₂¹) + 2(||f'||₂² + ||g'||₂²)

= (2-11/1²)(||f||₂¹ + ||g||₂¹)

= 3(||f||₂¹ + ||g||₂¹)

= 3(p(f) + p(g)),

where we have used the Cauchy-Schwarz inequality and the fact that ||f+g||₂ ≤ ||f||₂ + ||g||₂. Therefore, p(f) satisfies the triangle inequality.

In summary, we have shown that p(f) satisfies all the necessary properties to be a norm. Now, let's examine whether it is equivalent to the standard L²-norm ||f||₂.

We say that two norms p and q on a vector space V are equivalent if there exist positive constants A and B such that for all v ∈ V, Aq(v) ≤ p(v) ≤ Bq(v).

Let's first consider the upper bound. For any f ∈ L²[0, 1], we have:

p(f) = - (11/1²) + ||f||₂¹ + ∫₀¹ |f'|² dx

≤ 1||f||₂ + ||f'||₂² + 1

≤ (√2 + 1) ||f||₂,

where we have used the Cauchy-Schwarz inequality and the fact that ||f'||₂² ≤ 2||f||₂². Therefore, p is dominated by ||·||₂ with a constant of √2 + 1.

Now let's consider the lower bound. For any ε > 0, we can choose a function f ∈ L²[0, 1] such that:

(11/1²) + ||f||₂¹ + ∫₀¹ |f'|² dx < ||f||₂ + ε

This is possible because ||f'||₂ → 0 as ||f||₂ → 0. Therefore, p cannot be bounded below by any multiple of ||·||₂, since ||f||₂ can be

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It is impossible to determine the relationship between x and y from the question about the triangle. Hence,  the correct option is D. This is because none of the sides are labelled x or y from the information provided.

It is important to note that:

The side with the highest length is the side that is opposite the biggest angle.

It is also important to note that the angle that is least will always be opposite the side that is smallest.

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One pair of vertical angle are ∠1 and ∠4 ; ∠2 and  ∠3

One pair of adjacent angles are ∠1 and ∠2; ∠3 and ∠4; ∠1 and ∠3; ∠2 and ∠4.

One pair of supplementary angles are ∠1 and ∠2; ∠3 and ∠4.

What are vertical angles?

When two lines intersect, there are two vertical angles.

The vertical opposite angle of ∠1 is ∠4.

The vertical opposite angle of ∠2 is ∠3.

The vertical angles are ∠1 and ∠4; ∠2 and ∠3.

When two angles have a similar vertex and side, they are referred to as neighboring angles. The vertex of an angle is the point at which the rays that make up its sides come to an end.

∠1 and ∠2 have similar vertex and side. ∠2 and ∠3 have similar vertex and side. ∠3 and ∠4 have similar vertex and side. ∠3 and ∠4 have similar vertex and side.

Angles that add up to 180 degrees are referred to as supplementary angles.

∠1 and ∠2 is linear pair, it supplementary angles.

∠2 and ∠4 is linear pair, it supplementary angles.

∠3 and ∠4 is linear pair, it supplementary angles.

∠1 and ∠3 is linear pair, it supplementary angles.

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\(iii\implies i^3\implies i^2\cdot i\implies (-1)\cdot \sqrt{-1}\implies -\sqrt{-1}\implies -i\)

Answer:

(5/2,5)

Step-by-step explanation:

Answer and Step-by-step explanation:

Everything is in the picture.

Lines are forever, so they have an arrow at each end.

Line segments have a dot at each end, so they don't go on forever.

A ray has a dot on one end and an arrow on the other, so one side goes on forever.

An angle can be acute, right, or obtuse.

#teamtrees #PAW (Plant And Water)

Answer:

Step-by-step explanation:

17)

It's a 6-sided shape (hexagon) so we do 360 / 6 = 60

x = 60

18)

There are 540 total degrees in a pentagon

We have 82 + 36 + 70 = 188

540 - 188 = 352

x and the other x are equal, so

x = 188 / 2 = 94

x = 94

Whenever a set X is a subset of set Y, we say the Y is a superset of X and we write, Y ⊇ AX.

Symbol ⊇ is used to denote ‘is a super set of’

The correct answer is the janitor is mopping at a rate of 1/15 of a classroom per minute and this is a proper fraction.

It is given in the question that a school janitor has moped \(\frac{1}{3}\) of classroom in 5 minutes.

We have to find the rate at which he is moping

We will divide \(\frac{1}{3}\) by 5, to divide by 5 we will multiply \(\frac{1}{3}\) with  reciprocal of 5 that is \(\frac{1}{5}\)

\(\frac{1}{3}\)× \(\frac{1}{5}\) = \(\frac{1}{15}\)

Hence, the janitor is mopping at a rate of 1/15 of a classroom per minute.

This fraction is proper fraction If the numerator of a fraction is less than the denominator, the fraction is said to be a correct fraction.

A proper fraction's value is never more than 1. Sam, for instance, divided a chocolate bar into four equal pieces. He took one portion and handed his sister Rachel three portions.

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