pleaseeeee help me with this algebra question! select the solutions for the quadratic equation!

If a kite is flying 95 ft. off the ground, and its string is pulled taut. The angle of elevation of the kite is 59 degrees. Then the length of the string will be 110.8 ft.

Given information constitutes the following,

The distance of the flying kite from the ground, length AB (refer the figure) = 95 ft.

The angle of elevation of the kite, ∠ACB = 59°

We have to find the length of the string, that is the length AC. For that, we can apply Trigonometry as shown in the next steps of the solution.

In ΔABC, as shown in the attached figure,

sin (∠ACB ) = AB / AC

⇒ sin (59°) = 95 / AC

0.8572 = 95 / AC

AC = 95 / 0.8572

AC = 110.814

AC ≈ 110.8 ft.                [After rounding off to the nearest tenth]

Hence, the length of the string comes out to be 110.8 ft.

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The intersection of the two intervals in the number line will be = 4, 5,  6,.......+∞ = (-∞,6).

The union of the two intervals = -3, -2, -1, 0, 1, 2, 3, 4, 5,.....,+∞ = (-∞),9)

The given intervals are;

First interval  = (-∞, 6)

Second interval  = (-∞, 9)

Using the number line, we therefore, the first interval includes, -3, -2, -1, 0, 1, 2, 3, 4, 5,.....,+∞

The second interval includes, 4, 5,.....,+∞

Which gives the intersection as 4, 5, 6,7, 8,9......+∞

The union is the interval that combines the two sets of intervals which is given as follows;

The union of the two intervals = -3, -2, -1, 0, 1, 2, 3, 4, 5,.....,+∞

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The exponential function is illustrated below.

An exponential function has a growth factor or 3.76. What is the percentage growth rate?

The growth factor (b) is given as:

b = 3.76

So, the percentage growth rate (r) is calculated as:

r = b - 1

Substitute known values

r = 3.76 - 1

Evaluate the difference

r = 276%

The way to solve Financial Problems using Sequences & Series will be:

The first salary that Mr James earn is 10000 and there is a yearly increase of 2000. Find his salary in the 5th year. This will be:

= a + (n - 1)d

= 1000 + (5 - 1)2000

= 10000 + (4 × 2000)

= 10000 + 8000.

= 18000

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Using proportions, it is found that option A gives a figure that is a scale drawing of the same courtyard using the scale 1 centimeter = 3 feet.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

Researching this problem on the internet, the figure with a scale of 1 cm = 4 feet has the dimensions of:

51 cm, 75 cm, 30 cm and 72cm.

For a scale of 1 centimeter = 3 feet, these measures will be multiplied by 4/3, hence the figure is given in option A, as:

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The frequency distribution based on the information given is illustrated below.

A frequency distribution table is the

chart that summarizes all the data under two columns - variables/categories, and their frequency.

It should be noted that the distribution table has two or three columns and the first column lists all the outcomes as individual values or in the form of class intervals, depending upon the size of the data set.

Given the above information the frequency distribution table is:

Cost of living Number of states

85.0 - 94.9 9

95.0 - 104.9 5

105.0 - 114.9 0

115.0 - 124.9 2

125.0 - 134.9 2

135.0 - 144.9 1

145.0 - 154.9 1

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Answer:

Step-by-step explanation:

in the smallest triangle (BCD) you have an angle of 90° and one of 63°, the sum of the internal angles in a triangle is 180°, remove the known angles from 180 ° and you will have the measure of the CBD angle

180 - 63 - 90 =

27°

Answer:

27°

Step-by-step explanation:

180°-90°-63° = 27°

The period of [tex]f(x)[/tex] is [tex]\boxed{2\pi}[/tex].

Recall that [tex]\sin(x)[/tex] and [tex]\cos(x)[/tex] both have periods of [tex]2\pi[/tex]. This means

[tex]\sin(x + 2\pi) = \sin(x)[/tex]

[tex]\cos(x + 2\pi) = \cos(x)[/tex]

Replacing [tex]x[/tex] with [tex]3x[/tex], we have

[tex]\sin(3x + 2\pi) = \sin\left(3 \left(x + \dfrac{2\pi}3\right)\right) = \sin(3x)[/tex]

In other words, if we change [tex]x[/tex] by some multiple of [tex]\frac{2\pi}3[/tex], we end up with the same output. So [tex]\sin(3x)[/tex] has period [tex]\frac{2\pi}3[/tex].

Similarly, [tex]\cos(5x)[/tex] has a period of [tex]\frac{2\pi}5[/tex],

[tex]\cos(5x + 2\pi) = \cos\left(5 \left(x + \dfrac{2\pi}5\right)\right) = \cos(5x)[/tex]

We want to find the period [tex]p[/tex] of [tex]f(x)[/tex], such that

[tex]f(x + p) = f(x)[/tex]

[tex] \implies \sin(3x + p) + \cos(5x + p) = \sin(3x) + \cos(5x)[/tex]

On the left side, we have

[tex]\sin(3x + p) = \sin(3x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \sin(3x+2\pi) \cos(p-2\pi) + \cos(3x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \sin(3x) \cos(p-2\pi) + \cos(3x) \sin(p - 2\pi)[/tex]

and

[tex]\cos(5x + p) = \cos(5x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \cos(5x+2\pi) \cos(p-2\pi) - \sin(5x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \cos(5x) \cos(p-2\pi) - \sin(5x) \sin(p-2\pi)[/tex]

So, in terms of its period, we have

[tex]f(x) = \sin(3x) \cos(p - 2\pi) + \cos(3x) \sin(p - 2\pi) \\\\ ~~~~~~~~ ~~~~+ \cos(5x) \cos(p - 2\pi) - \sin(5x) \sin(p - 2\pi)[/tex]

and we need to find the smallest positive [tex]p[/tex] such that

[tex]\begin{cases} \cos(p - 2\pi) = 1 \\ \sin(p - 2\pi) = 0 \end{cases}[/tex]

which points to [tex]p=2\pi[/tex], since

[tex]\cos(2\pi-2\pi) = \cos(0) = 1[/tex]

[tex]\sin(2\pi - 2\pi) = \sin(0) = 0[/tex]

f(x) = x³ + 2x² - 15x - 36 is the equation of a degree 3 polynomial (in factored form) with the given zeros of f(x) are − 3 , 4 , − 3 assuming that the leading coefficient is 1. This can be obtained by formula of polynomial function.

⇒ f(x) =  a(x - r₁)(x - r₂)(x - r₃)

where a is the leading coefficient

Here in the question it is given that,

By using the formula of polynomial function we get,

⇒ f(x) =  a(x - r₁)(x - r₂)(x - r₃)

⇒ f(x) = 1(x - (-3))(x - (4))(x - (-3))

⇒ f(x) = 1(x + 3)(x - 4)(x + 3)

⇒ f(x) = (x + 3)(x² - x - 12)

⇒ f(x) = x³ - x² - 12x + 3x² - 3x - 36

⇒ f(x) = x³ + 2x² - 15x - 36

Hence f(x) = x³ + 2x² - 15x - 36 is the equation of a degree 3 polynomial (in factored form) with the given zeros of f(x): − 3 , 4 , − 3 assuming that the leading coefficient is 1.

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

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

Take HJ = a, GH = b and GJ = c

put the value of a from equation 1 in equation 2

[tex]\qquad❖ \: \sf \:c = (b + 2) + b - 17[/tex]

[tex]\qquad❖ \: \sf \:c = 2b - 15[/tex]

now, put the value of a and c in equation 3

[tex]\qquad❖ \: \sf \:b + 2 + b + 2b - 15 = 73[/tex]

[tex]\qquad❖ \: \sf \:4b - 13 = 73[/tex]

[tex]\qquad❖ \: \sf \:4b = 86[/tex]

[tex]\qquad❖ \: \sf \:b = 21.5 \: \: in[/tex]

Now, we need to find HJ (a)

[tex]\qquad❖ \: \sf \:a = b + 2[/tex]

[tex]\qquad❖ \: \sf \:a = 21.5 + 2[/tex]

[tex]\qquad❖ \: \sf \:23.5 \: \: in[/tex]

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

Answer:

23.5 in

Step-by-step explanation:

To find the length of HJ in triangle GHJ, create three equations using the given information, then solve simultaneously.

Equation 1

HJ is two inches longer than GH:

⇒ HJ = GH + 2

Equation 2

GJ is 17 inches shorter than the sum of HJ and GH:

⇒ GJ + 17 = HJ + GH

Equation 3

The perimeter of ΔGHJ is 73 inches:

⇒ HJ + GH + GJ = 73

Substitute Equation 1 into Equation 2 and isolate GJ:

⇒ GJ + 17 = GH + 2 + GH

⇒ GJ + 17 = 2GH + 2

⇒ GJ = 2GH - 15

Substitute Equation 1 into Equation 3 and isolate GJ:

⇒ GH + 2 + GH + GJ = 73

⇒ 2GH + GJ = 71

⇒ GJ = 71 - 2GH

Equate the two equations where GJ is the subject and solve for GH:

⇒ 2GH - 15 = 71 - 2GH

⇒ 4GH = 86

⇒ GH = 21.5

Substitute the found value of GH into Equation 1 and solve for HJ:

⇒ HJ = 21.5 + 2

⇒ HJ = 23.5

If the length of hypotenuse is 95 m ,perpendicular is 57 m then the length of missing leg is 76m.

Given that the length of hypotenuse is 95 m ,the length of perpendicular is 57 m.

We are required to find the length of base or missing leg.

The given triangle is a right angled triangle. We can easily find out the length of the base of the triangle by using pythagoras theorem.

Pythagoras theorem says that the square of hypotenuse of a right angled triangle is equal to the sum of squares of the base and perpendicular of that triangle.

[tex]H^{2} =P^{2} +B^{2}[/tex]

We have to find the base of the triangle.

B=[tex]\sqrt{H^{2} -P^{2} }[/tex]

=[tex]\sqrt{(95)^{2} -(57)^{2} }[/tex]

=[tex]\sqrt{9025-3249}[/tex]

=[tex]\sqrt{5776}[/tex]

=76 m.

Hence if the length of hypotenuse is 95 m ,perpendicular is 57 m then the length of missing leg is 76m.

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The measure of angle 1 (m ∠1) is 28° and the measure of angle 2 (m ∠2) is 62°

From the question, we are to determine the measure of angle 1 and the measure of angle 2

The given diagram is a kite and the diagonals intersect at right angles

Thus,

m ∠2 + 28° + 90° = 180°

m ∠2 = 180° - 28° - 90°

m ∠2 = 62°

Hence, the measure of angle 2 is 62°

For the measure of angle 1

Consider ΔADB

ΔADB is an isosceles triangle

Thus,

In the triangle, m ∠D = m ∠B

Then, we can write that

m ∠1 + 62° + 90° = 180°

m ∠1 = 180° - 62° - 90°

m ∠1 = 28°

∴ The measure of angle 1 is 28°

Hence, the measure of angle 1 (m ∠1) is 28° and the measure of angle 2 (m ∠2) is 62°

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Answer:

50°

Step-by-step explanation:

Note EFGH is an isosceles trapezoid.

∠HGF=77° (base angles of an isosceles trapezoid are congruent)

∠EGH=27° (angles in a triangle add to 180°)

∠FGE=50° (angle subtraction postulate)

The probability that the student plays football is 20/33.

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

The more likely the event is to happen, the closer the probability value would be to 1. The less likely it is for the event not to happen, the closer the probability value would be to zero.

The probability that the student plays football = total number of students who play football / total number of students

The probability that the student plays football = 40/66 = 20/33

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13) 120 + 1.5 + 0.25 = 121.75 pounds

14) 4 - 1.5 = 2.5 pints

15) (2.25)(32)= $72

As per the unitary method, Mrs. Smith's current weight is 121 pounds and 3 ounces.

To find Mrs. Smith's current weight, we need to add the weight she gained over the last two months to her initial weight. First, we will convert the mixed fractions to improper fractions for easier calculations.

1½ pounds can be written as (2 * 1) + 1/2 = 3/2 pounds.

1/4 pound remains as it is.

Now, let's add the weight gained in the last two months:

3/2 pounds + 1/4 pound = (3/2) + (1/4) = (6/4) + (1/4) = 7/4 pounds.

Next, we add the total weight gained to Mrs. Smith's initial weight:

120 pounds + 7/4 pounds = (120 * 4/4) + (7/4) = (480/4) + (7/4) = 487/4 pounds.

To express the answer in pounds, we convert the improper fraction back to a mixed fraction:

487/4 pounds can be written as (4 * 121) + 3 pounds.

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Due to length restrictions, we kindly invite to check the explanation herein for further details of the hyperbola.

Herein we have an hyperbola whose axis of symmetry is parallel to the y-axis and the major semiaxis length is in the y-direction. By analytical geometry, we know that eccentricities of hyperbolae are greater than 1.

a) The formula for eccentricity is:

e = √(a² + b²) / a      (1)

Where:

If we know that a = 4 and b = 3, then the eccentricity of the hyperbola is:

e = √(4² + 3²) / 4

e = 5 / 4

b) The coordinates of the two vertices of the hyperbola are:

V(x, y) = (h, k ± a)      (2)

Where (h, k) are the coordinates of the center of the hyperbola.

V₁ (x, y) = (0, 4), V₂ (x, y) = (0, - 4)

The coordinates of the foci of the hyperbola are:

F(x, y) = (h, k ± c), where c = √(a² + b²).     (3)

c = √(4² + 3²)

c = 5

F₁ (x, y) = (0, 5), F₂ (x, y) = (0, - 5)

The equations of the asymptotes of the hyperbola are:

y = ± (a / b) · x

y = ± (4 / 3) · x      (4)

And the equations of the directrices of the hyperbola are:

y = k ± (2 · a - c)

y = 0 ± (8 - 5)

y = ± 3     (5)

The graph is presented in the image attached below.

c) The parametric equations for the hyperbola are the following formulae:

y = ± a · cosh t   →   y = ± 4 · cosh t      (6)

x = b · sinh t   →   x = 3 · sinh t     (7)

d) First, we determine the slopes of the two tangent lines by implicit differentiation:

m = (16 · x) / (9 · y)

m = (16 · 2.3) / [9 · (± 4.807)]

m = ± 0.851

Second, we find the intercept of each tangent line:

(x, y) = (2, 4.807)

b = 4.807 - 0.851 · 2

b = 3.105

y = 0.851 · x + 3.105      (8)

(x, y) = (2, - 4.807)

b = - 4.807 - (- 0.851) · 2

b = - 3.105

y = - 0.851 · x - 3.105      (9)

e) The definite integral of the arc length of the hyperbola is presented below:

[tex]s = \int\limits^{2}_{1} {\sqrt{\left(\frac{dx}{dt} \right)^{2}+\left(\frac{dy}{dt} \right)^{2}}} \, dt[/tex]

If we know that dx / dt = a² · sinh² t and dy / dt = b² · cosh² t, then the definite integral for the arc length is:

[tex]s = \int\limits^2_1 {\sqrt{a^{2}\cdot \sinh ^{2}t +b^{2}\cdot \cosh^{2}t}} \, dt[/tex]      (10)

f) We apply the following substitutions on (1): x = r · cos θ, y = r · sin θ. Then, we have the polar form by algebraic handling:

r(θ) = (a · b) / (b² · sin² θ - a² · cos² θ)     (11)

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Step-by-step explanation:

Use this sort of layout, but where x will be an odd number, do not shade it. there should be a pattern of shaded segments followed by unshaded segments repeating

The limit does not exist. There are infinitely many infinite discontinuities at [tex]x=n\pi[/tex], where [tex]n\in\Bbb N[/tex]. The function oscillates wildly between negative and positive infinity.

The answers to the question are:

1. The letters of the word track can be arranged in 5! ways

These are  5 x 4 x 3 x 2 x1

= 120

2. The way that the student would be able to select 6 out of 10 questions would be by 10C6

= 210 ways

C)This teacher would be able to make the decision of the prices to the students using =20C6= n!(n-r!r!)

= 38760

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the area of the minor segment is 0. 29 cm^2

From the information given, we have the following parameters;

It is important to note the formula for area of a sector is given as;

Area = πr² + θ/360° - 1/ 2 r² sin θ

The value for π = 3.142

θ = 30°

Now, let's substitute the values

Area = 3. 142 × 5² × 30/ 360 - 1/ 2 × 5² × sin 30

Find the difference

Area = 3. 142 × 25 × 1/ 12 - 1/ 2 × 25 × 1/2

Multiply through

Area = 6. 54 - 6. 25

Area = 0. 29 cm^2

The area of the minor segment is given as 0. 29 cm^2

Thus, the area of the minor segment is 0. 29 cm^2

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Answer:

The shortest piece is the first piece and it is 14 cm long.

Step-by-step explanation:

We have three unknowns so we need 3 equations.

Let x = the length of the first piece

Let y = the length of the second piece

Let z = the length of the third piece.

x + y + z = 74            y = 2x          z = y + 4

There are a number of ways to solve this.  I am going to plug in 2x for y into the first and the third equation to get:

x + y + z = 74

x + 2x + z = 74 Combine the x terms

3x + z = 74

Next, I am going to substitute 2x in for y in the third equation above.

z = y + 4

z = 2x + 4  I am going to put both variable on the left side of the equation

z - 2x = 4

I can know take the two bold equations that I have above and solve for the either x or z.  I am going to solve for z.  I need one of the equation to have a z and the other equation to have -z so that they will cancel one another out.  I am going to multiple z - 2x = 4 all the way through by -1 to get:

z - 2x = 4

-1(z - 2x) = 4(-1)

-z +2x = -4  

I am going to rearrange 3x + z = 74 so that the z term is first and add it to -z + 2x = -4

z + 3x = 74

-z + 2x = -4

      5x = 70 divide both sides by 5

x = 14  This is the length of the first piece.

y = 2x

y = 2(14) = 28

y = 28 This is the length of the second piece.

z = y+4

z = 28 + 4 = 32

or

x + y + z = 74

14 + 28 + z = 74

42 + z = 74  Subtract 42 from both sides.

z = 32

The system of inequalities y < -x+4 and y >-x 2 do not have a solution

Inequalities are expressions that have unequal values when compared or evaluated

The system of inequalities is given as

y < -x+4

y >-x 2

Next, we plot the inequalities on a graphing tool

See attachment for the graph

From the attached graph, the lines of the inequalities do not intersect

This means that the system of inequalities do not have a solution

Hence, the system of inequalities y < -x+4 and y >-x 2 do not have a solution

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By critically observing the number lines, the intersection of both (-∞, 6) and (-∞, 9) is (6, 9) because this is the point where they overlap. Also, the union of both (-∞, 6) and (-∞, 9) on a number line is (-∞, 9).

A number line can be defined as a type of graph with a graduated straight line which contains both positive and negative numerical values that are placed at equal intervals along its length.

Given the following intervals:

On a number line, the first interval would comprise the following numerical values -∞,..........-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6.

On a number line, the second interval would comprise the following numerical values -∞,..........-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

By critically observing the number lines, we can logically deduce that intersection of both (-∞, 6) and (-∞, 9) is (6, 9) because this is the point where they overlap.

Also, the union of both (-∞, 6) and (-∞, 9) on a number line is (-∞, 9).

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Answer:

[tex]\sf 28 \frac{1}{3} \:ft=28.3\:ft\:(nearest\:tenth)[/tex]

Step-by-step explanation:

Given information:

Let x be the unknown length of the other side of Mr Kelly's house.

To solve, set up a ratio with the given information and the defined unknown, then solve for x:

[tex]\textsf{20 ft : 6 strides = x ft : 8.5 strides}[/tex]

[tex]\implies \sf 20:6 = x:8.5[/tex]

[tex]\implies \sf \dfrac{20}{6}=\dfrac{x}{8.5}[/tex]

[tex]\implies \sf x=\dfrac{20 \cdot 8.5}{6}[/tex]

[tex]\implies \sf x=\dfrac{170}{6}[/tex]

[tex]\implies \sf x=28 \frac{1}{3} \:ft[/tex]

[tex]\implies \sf x=28.3\:ft\:(nearest\:tenth)[/tex]

Therefore, the length of the other side of Mr Kelly's house that takes him 8.5 strides to walk is 28.3 ft (nearest tenth).

Let that be x

Since [tex]a,b,c[/tex] are in geometric progression, if [tex]r[/tex] is the common ratio between consecutive terms, then

[tex]a=a[/tex]

[tex]b = ar[/tex]

[tex]c=ar^2[/tex]

Since [tex]a,b,c[/tex] are also in arithmetic progression, if [tex]d[/tex] is the common difference between consecutive terms, then

[tex]a = a[/tex]

[tex]b = a + d \implies d = b-a[/tex]

[tex]c = b + d = a + 2d \implies c = a + 2(b-a) = 2b-a[/tex]

Given that [tex]abc=27[/tex], we have

[tex]abc = a\cdot ar\cdot ar^2 = (ar)^3 = 27 \implies ar = 3 \implies a = \dfrac3r[/tex]

[tex]b = \dfrac3r \cdot r = 3[/tex]

[tex]c = \dfrac3r \cdot r^2 = 3r[/tex]

It follows that

[tex]c = 2b-a \iff 3r = 6 - \dfrac3r[/tex]

Solve for [tex]r[/tex].

[tex]3r - 6 + \dfrac3r = 0[/tex]

[tex]3r^2 - 6r + 3 = 0[/tex]

[tex]r^2 - 2r + 1 = 0[/tex]

[tex](r-1)^2 = 0[/tex]

[tex]\implies r=1 \implies a=b=c=3[/tex]

so the only possible sequence is {3, 3, 3, …}.

The simplification of a³ - 1000b³ and 64a³ - 125b³ is (a - 10b) × (a² + 10ab + 100b²) and 4a - 5b) • (16a² + 20ab + 25b²) respectively.

Question 1: a³ - 1000b³

a³ - b³

= (a-b) × (a² +ab +b²)

So,

(a - 10b) × (a² + 10ab + 100b²)

Question 2: 64a³ - 125b³

a³ - b³

= (a-b) × (a² +ab +b²)

So,

(4a - 5b) • (16a² + 20ab + 25b²)

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Answer:

15u+15x-20

Step-by-step explanation:

* = multiply or times

We have to multiply everything in the parenthesis by -5, meaning:

-5*-3u = 15u

-5*-3x = 15x

-5*4 = -20

Put it all together: 15u+15x-20

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{What is the distributive formula?}[/tex]

[tex]\mathsf{a(b + c)}\\\\\mathsf{= a(b) + a(c)}\\\\\mathsf{= ab + ac}[/tex]

[tex]\large\textbf{Extended distributive property formula:}[/tex]

[tex]\mathsf{a(b + c + d)}\\\\\mathsf{= a(b) + a(c) + a(d)}\\\\\mathsf{= ab + ac + ad}[/tex]

[tex]\huge\textbf{What are we solving for?}[/tex]

[tex]\large\textbf{It seems like you have the extended distributive property}[/tex]

[tex]\mathsf{-5(-3u - 3x + 4)}[/tex]

[tex]\huge\textbf{What are the steps to solving for the}\\\\\huge\textbf{given equation?}[/tex]

[tex]\mathsf{-5(-3u - 3x + 4)}\\\\\mathsf{= -5(-3u) - 5(-3x) - 5(4)}\\\\\mathsf{= 15u + 15x - 20}[/tex]

[tex]\huge\textbf{What is the answer to the equation?}[/tex]

[tex]\huge\boxed{\frak{{= 15\mathsf{u} + 15\mathsf{x} - 20}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

Answer:

[tex]h = \bf 28.3 \space\ m[/tex]

Step-by-step explanation:

• We are given:

○ Volume = 36 m³,

○ Circumference = 4 m

• Let's find the radius of the cylinder first:

[tex]\mathrm{Circumference} = 2 \pi r[/tex]

Solving for [tex]r[/tex] :

⇒ [tex]4 = 2 \pi r[/tex]

⇒ [tex]r = \frac{4}{2\pi}[/tex]

⇒ [tex]r = \bf \frac{2}{\pi}[/tex]

• Now we can calculate the height using the formula for volume of a cylinder:

[tex]\mathrm{Volume} = \boxed{\pi r^2 h}[/tex]

Solving for [tex]h[/tex] :

⇒ [tex]36 = \pi \cdot (\frac{2}{\pi}) ^2 \cdot h[/tex]

⇒ [tex]h = \frac{36 \pi^2}{4 \pi}[/tex]

⇒ [tex]h = 9 \pi[/tex]

⇒ [tex]h = \bf 28.3 \space\ m[/tex]

Answer:

9π m ≈ 28.27m

Step-by-step explanation:

The volume of a right cylinder is given by the formula

πr²h where r is the radius of the base of the cylinder(which is a circle), h is the height of the cylinder

Circumference of base of cylinder is given by the formula 2πr

Given,

2πr = 4m

r = 2/π m

Volume given as 36 m³

So πr²h = 36
π (2/π)² h = 36

π x 4/π² h = 36

(4/π) h = 36

h = 36π/4 = 9π ≈ 28.27m

The value of the composite functions (g∘f)(x) and (f∘g)(x)   are 32x^2 + 16x + 3 and 8x^2 + 5 respectively

Composite function is also known as function of a function. They are determined by representing x with the other function.

Given the following functions

f(x)=4x+1

g(x)=2x^2+1

(f∘g)(x) = f(g(x))

(f∘g)(x) = f(2x^2+1)

(f∘g)(x) = 4(2x^2+1) + 1

(f∘g)(x)  =8x^2 + 5

For the composite function (g∘f)(x)

(g∘f)(x) = g(f(x))

(g∘f)(x) = g(4x+1)

Replace x wit 4x+1  to have:

(g∘f)(x) = 2(4x+1)^2 + 1

(g∘f)(x)= 2(16x^2+8x+1) + 1

(g∘f)(x) = 32x^2 + 16x + 3

Hence the value of the composite functions (g∘f)(x) and (f∘g)(x)   are 32x^2 + 16x + 3 and 8x^2 + 5 respectively

Learn more on composite function here: https://brainly.com/question/10687170

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