Find f(4). what does f(4) represent

Answer:

a)  The angle subtended by the chord at the centre of the circle is 120°.

b)  The length of the minor arc is 12.57 cm (2 d.p.).

The perimeter of the minor segment formed by a chord and minor​ arc is 22.96 cm (2 d.p.).

Step-by-step explanation:

The chord of a circle is the base of an isosceles triangle, where its congruent sides are the radii of the circle.

The height of the isosceles triangle is the perpendicular bisector of the chord from the central angle.

The angle subtended by the chord at the centre of the circle is twice the angle formed by the radius and the perpendicular bisector (marked θ on the attached diagram).

To find the measure of angle θ, use the cosine trigonometric ratio.

[tex]\begin{aligned}\implies \cos \theta&=\sf \dfrac{adjacent\;side}{hypotenuse}\\\\\cos \theta &=\sf \dfrac{3}{6}\\\\\cos \theta &=\sf \dfrac{1}{2}\\\\\theta &=\arccos\left(\sf \dfrac{1}{2}\right)\\\\\theta &=60^{\circ}\end{aligned}[/tex]

As the angle subtended by the chord at the centre of the circle is twice angle θ:

[tex]\implies 2\theta=2 \cdot 60^{\circ}=120^{\circ}[/tex]

Therefore, the angle subtended by the chord at the centre of the circle is 120°.

A minor arc is less than 180° and is equal to the central angle.

Therefore, as the angle subtended by the chord at the centre of the circle is 120°, the minor arc is the part of the circumference between the two endpoints of the chord (marked in red on the attached diagram).

To find the length of the minor arc, use the arc length formula.

[tex]\boxed{\begin{minipage}{6.4 cm}\underline{Arc length}\\\\Arc length $= \dfrac{\pi r\theta}{180^{\circ}}$\\\\where:\\ \phantom{ww}$\bullet$ $r$ is the radius. \\ \phantom{ww}$\bullet$ $\theta$ is the angle measured in degrees.\\\end{minipage}}[/tex]

Therefore:

[tex]\begin{aligned}\implies \textsf{Arc length}&= \dfrac{\pi \cdot 6 \cdot 120^{\circ}}{180^{\circ}}\\\\&= 4 \pi\\\\&= 12.57\; \sf cm\;(2\;d.p.)\end{aligned}[/tex]

Therefore, the length of the minor arc is 12.57 cm (2 d.p.).

To calculate the perimeter of the minor segment formed by the chord and minor​ arc, we need to calculate the length of the chord.

[tex]\boxed{\begin{minipage}{10.2 cm}\underline{Chord length}\\\\Chord length $=2r\sin\left(\dfrac{\theta}{2}\right)$\\\\where:\\ \phantom{ww}$\bullet$ $r$ is the radius. \\ \phantom{ww}$\bullet$ $\theta$ is the angle (in degrees) subtended at the center by the chord.\\\end{minipage}}[/tex]

Therefore:

[tex]\begin{aligned}\implies \textsf{Chord length}&= 2 \cdot 6 \cdot \sin \left(\dfrac{120^{\circ}}{2}\right)\\\\&= 12 \cdot \sin \left(60^{\circ}\right)\\\\&= 12 \cdot \dfrac{\sqrt{3}}{2}\\\\&= 6\sqrt{3}\; \sf cm\end{aligned}[/tex]

Therefore, the perimeter of the minor segment formed by the chord and minor​ arc is:

[tex]\begin{aligned}\implies \textsf{Perimeter}&=\textsf{Minor arc length}+\textsf{Chord length}\\&=4 \pi + 6 \sqrt{3}\\&=22.96\; \sf cm\;(2\;d.p.)\end{aligned}[/tex]

As we see, triangle ABC is congruent to triangle BMK so, a) length of AC = 5

b) length of BM = 4.5

c) length of KM = 3

d) length of BM = 3.42 , MC = 17.58

We have a triangle ABC and KM, a segment is parallel to AC, i.e., KM || AC. First we show that triangle ABC is congruent to triangle BMK. AC and AB are transversals that intersect parallel lines. Since the two pairs of corresponding angles created by this intersection are equal, that is,

∠ BKM =∠ BAC and ∠ BMK = ∠ BCA

By Angle-Angle congruence rule, ∆ ABC is similar to triangle BMK, i.e., ∆ABC ~ ∆BMK. When two triangles are similar, the ratios of the lengths of their corresponding sides are equal, KB/AB = KM/AC --(1) or

BM/BC = KM/AC --(2)

a) AB = 10, KB = 2, KM = 1, determine AC=?

from equation (1), KB/AB = KM/AC

=> 2/10 = 1/AC

Cross multiplication

=> 2 AC = 10

=> AC = 5

b) KM= 3, AC = 6,BC = 9, determine BM= ?

from equation (2), BM/BC = KM/AC

=> BM/9 = 3/6

Cross multiplication

=> 6 BM = 27

=> BM = 9/2

=> BM = 4.5

c)BC = 25, MC = 10, AC = 5, determine KM= ?

BM = BC - MC = 25 - 10 = 15

from equation (2), BM/BC = KM/AC

=> 15/25 = KM/5

Cross multiplication

=> 25 KM = 15×5

=> KM = 75/25

=> KM = 3

d)AK = 10, KB = 4, BC = 21, determine BM=? and MC= ?

AB = AK + KB = 10 + 4 = 14

from equations (1) and (2),BK/AB = BM/BC

=> 4/14 = BM/21

Cross multiplication

=> 14 BM = 21 × 4

=> BM = 84/14 = 24/7

=> BM = 3.42

so, MC = BC - BM = 21 - 3.42 = 17.58

Hence, required value is 3.42..

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Complete question:

Given: ∆ABC, KM|| AC. See the above figure.1.

a) AB=10, KB=2, KM=1

AC-?

b) KM=3, AC=6,BC=9

BM-?

c)BC=25, MC=10, AC=5

KM-?

d)AK=10,KB=4,BC=21

BM-?,MC-?

I. Simplify the following fully:

1.1 (2x-1)(x^(2)-3x+1)

= 2x^(3) - 6x^(2) + 2x - x^(2) + 3x - 1

= 2x^(3) - 7x^(2) + 5x - 1

1.2 12(2x+3)(5-x)

= 12(-2x^(2) + 7x + 15)

= -24x^(2) + 84x + 180

1.3 (x^(3)-1)/((x+2)+x(x+2))+(x-1)/(2x+4)

= (x^(3)-1)/(x^(2)+3x+2)+(x-1)/(2(x+2))

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

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

= (2x^(3)+x^(2)-x-1+x^(2)-1)/(2(x+2)(x+1))

= (2x^(3)+2x^(2)-x-2)/(2(x+2)(x+1))

1.4 (3)/(x^(2)-9)+(2)/((x-3)^(2))

= (3)/((x+3)(x-3))+(2)/((x-3)^(2))

= (3(x-3)+2(x+3))/(x+3)(x-3)^(2)

= (3x-9+2x+6)/(x+3)(x-3)^(2)

= (5x-3)/(x+3)(x-3)^(2)

2. Factorise:

2.1 6x^(2)+7x-20

= (2x-4)(3x+5)

2.2 x^(3)+x^(2)-x-1

= (x+1)(x^(2)-1)

= (x+1)(x+1)(x-1)

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In the triangle DEF the measure of angle E is obtained as (183 - 9x)°.

What are triangles?

Triangles are a particular sort of polygon in geometry that have three sides and three vertices. Three straight sides make up the two-dimensional figure shown here. An example of a 3-sided polygon is a triangle. The total of a triangle's three angles equals 180 degrees. One plane completely encloses the triangle.

In the triangle DEF the value for ∠D = (4x + 1)°.

In the triangle DEF the value for ∠F = (5x - 4)°.

According to the properties of a triangle, the sum of angles of a triangle is equal to 180°.

So, the equation will be -

(4x + 1)° + (5x - 4)° + ∠E = 180°

Evaluate the expression -

(9x - 3)° + ∠E = 180°

∠E = 180° - (9x - 3)°

∠E = 180° - 9x + 3°

∠E = (183 - 9x)°

Therefore, the value for angle E is (183 - 9x)°.

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v1, v2, and v3 are linearly independent

Yes, the vectors v1, v2, and v3 are linearly independent.

To verify this, we can set up the equation:
a1v1 + a2v2 + a3v3 = 0

If the vectors are linearly independent, then the only way to solve this equation is if a1 = a2 = a3 = 0.

Substituting the values for v1, v2, and v3 into the equation yields:

-123 - 3a1 + 16 - 4a2 - 4a3 - 1640a3 = 0

Solving for a1, a2, and a3, we get:

a1 = 0, a2 = 0, a3 = 0

Therefore, the vectors v1, v2, and v3 are linearly independent.

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Answer:

37

Step-by-step explanation:

The simplified answer is 16y² - 25x².

When multiplying conjugate binomials,

We use the formula (a+b)(a-b) = a²-b².

In this case, a = 4y and b = 5x. So,

We can pluggin these values into the formula and simplify:

(4y+5x)(4y-5x) = (4y)² - (5x)²

= 16y² - 25x²

Therefore, the simplified answer is 16y² - 25x². This is the final answer, and there are no further steps to simplify.

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To find theta to the nearest tenth of a degree, we can use a calculator to find the inverse of the given trigonometric function. The inverse of a trigonometric function is denoted by a "-1" superscript.

For the first question, we have cos theta = 0.2442 with theta in QI. To find theta, we can use the inverse cosine function:
theta = cos^-1(0.2442) ≈ 75.7°

For the second question, we have cos theta = -0.7750 with theta in QII. To find theta, we can use the inverse cosine function:
theta = cos^-1(-0.7750) ≈ 139.2°
Since theta is in QII, we can subtract this angle from 180° to find the angle in QII:
theta = 180° - 139.2° ≈ 40.8°

For the third question, we have tan theta = 0.5860 with theta in QIII. To find theta, we can use the inverse tangent function:
theta = tan^-1(0.5860) ≈ 30.4°
Since theta is in QIII, we can add 180° to this angle to find the angle in QIII:
theta = 180° + 30.4° ≈ 210.4°

For the fourth question, we have cos theta = 0.2442 with theta in QI. This is the same as the first question, so the answer is the same:
theta = cos^-1(0.2442) ≈ 75.7°

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The graphs when classified are decay and growth functions

The complete solution is shown below

A graph that represents an exponential growth would increase as the input value (x) increases, while graphs of exponential decay would decrease with increment in x

This means that:

An exponential function is represented as

y = abˣ

If b > 1, then it is a growth function, otherwise it is a decay function

This means that:

For the equation y = abˣ, we have

Initial value = a

Decay rate (if b < 0) = 1 - b

Growth rate (if b > 0) = b - 1

This means that

Function (12) is unclear

13. f(x) = 11(0.4)ˣ:

Initial value = 11

Decay factor = 0.4

Decay rate = 6%

14. f(x) = (1/4)ˣ:

Initial value = 1

Decay factor = 1/4

Decay rate = 75%

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Any number that has the same ratiο as 1:2 and faIIs between 0 and 6 withοut repeating can be used fοr the divisiοns οf 2 by 4 and 3 by 6.

The basic unit οf mathematics is the number. Numbers are used fοr a variety οf activities, such as measuring, cοunting, and indexing. Accοrding tο their prοperties, numbers can be cIassified intο a variety οf categοries, incIuding naturaI numbers, whοIe numbers, ratiοnaI and irratiοnaI numbers, integers, reaI numbers, cοmpIex numbers, even and οdd numbers, etc. It is pοssibIe tο determine the οutcοme number using simpIe integer arithmetic οperatiοns.

In the given questiοn, we need tο find twο sets οf three distinct numbers between 0 and 6 that have the same ratiο.

One pοssibIe sοIutiοn is:

2 divided by 4.

= 3 divided by 6

Here:

The ratiο οf the first set is 2:4 οr 1:2. When we divide 2 by 4, we get 0.5.

The ratiο οf the secοnd set is 3:6 οr 1:2. When we divide 3 by 6, we aIsο get 0.5.

Therefοre, we have:

2 divided by 4.

= 0.5

3 divided by 6.

= 0.5

Sο, we can say:

2 divided by 4

= __ divided by __

where 3 and 6 can be any numbers that have the same ratiο as 1:2.SimiIarIy,

3 divided by 6

= __ divided by __

where 3 and 6 can be any numbers that have the same ratiο as 1:2.

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The points M, N, P, and Q have been plotted in the coordinate plane below.

In order to determine the midpoint of a line segment with two (2) endpoints, we would add each point together and divide by two (2).

Midpoint = [(x₁ + x₂)/2, (y₁ + y₂)/2]

Substituting the given parameters into the midpoint of a line segment formula, the midpoint of the diagonal is given by;

Midpoint MP = [(2 + 1)/2, (-1 + 2)/2]

Midpoint MP = [3/2, 1/2]

Midpoint MP = (1.5, 0.5).

For the fourth vertex Q, we have:

Midpoint = [(x₁ + x₂)/2, (y₁ + y₂)/2]

(1.5, 0.5) = [(4 + x₂)/2, (0 + y₂)/2]

4 + x₂ = 2(1.5)

x₂ = -4 + 3

x₂ = -1.

0 + y₂ = 2(0.5)

y₂ = 1.

Therefore, vertex Q = (-1, 1).

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The required real diameter οf the tin is 250 mm.

Diameter is a measurement οf the length οf a straight line that passes thrοugh the center οf a circular οbject, such as a circle οr a sphere. It is the distance between twο pοints οn the οbject's edge that are furthest frοm each οther and that pass thrοugh the center οf the οbject.

Here,

If the diameter οf the tin in the picture is 10 cm, then the actual diameter οf the tin can be calculated as:

Actual Diameter = 2.5 × Picture Diameter

Actual Diameter = 2.5 × 10 cm

Actual Diameter = 25 cm

Tο cοnvert the actual diameter tο millimeters (mm), we can multiply by 10:

Actual Diameter in mm = Actual Diameter × 10

Actual Diameter in mm = 25 cm × 10

Actual Diameter in mm = 250 mm

Therefοre, the real diameter οf the tin is 250 mm.

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Complete question:

There are 36 Snails present in the pond.

A branch of mathematics known as algebra deals with symbols and the mathematical operations performed on them.

Variables are the name given to these symbols because they lack set values.

In order to determine the values, these symbols are also subjected to various addition, subtraction, multiplication, and division arithmetic operations.

Given:

Ratio of Snail to Newts = 3:4

So, Number of Snail= 3x and number of Newts = 4x

If 12 more newts added then the ratio is 3:5

So, 3x/  4x+ 12 = 3/5

15 x = 3(4x+ 12)

15x = 12x + 36

15x - 12x = 36

3x = 36

x= 12

So, the number of Snails in pond are 3x = 3(12)= 36.

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The equation of the horizontal asymptote of P isy = 22818.18.  This means that the insect population will approach a maximum value of approximately 22818 insects, no matter how many months have passed since they were introduced. The unit of measure for this value is insects per month.

a. The equation of the horizontal asymptote can be found by taking the limit of P(t) as t approaches infinity.

lim P(t) as t approaches infinity = lim (50+251)/(2+0.011t) as t approaches infinity

Since the denominator grows without bound as t approaches infinity, the limit of P(t) approaches 251/0.011 = 22818.18. Therefore, the equation of the horizontal asymptote is:

y = 22818.18

b. The horizontal asymptote tells us the maximum population that the protected area can sustain for this type of insect. In other words, as time goes on, the insect population will approach a maximum value of approximately 22818 insects. The unit of measure for the asymptote value is insects.

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Answer: 48

Step-by-step explanation:

6 x 8 = 48

Answer:

Simplify both sides of the equation and combine all same-side like terms.

Combine opposite-side like terms to obtain the variable term on one side of the equal sign and the constant term on the other.

Divide or multiply as needed to isolate the variable.

The three basic operations used to simplify a linear system are elimination, substitution, and augmentation.

The 3 basic operations used to simplify a linear system are:
1. Swapping the order of equations: This operation is used to rearrange the equations in the system to make it easier to solve.
2. Multiplying or dividing an equation by a constant: This operation is used to make the coefficients of the variables in the equation equal to 1 or -1, which makes it easier to solve the system.
3. Adding or subtracting equations: This operation is used to eliminate one of the variables in the system, which reduces the number of equations and makes it easier to solve the system.
These operations are used to manipulate the equations in the system in order to find the values of the variables that satisfy all of the equations in the system. By using these operations, we can simplify the system and solve it more easily.

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The coordinate matrix [x]B , given the coordinate matrix [x]B'.
[x]B = P * [x]B' = [(4.67, -0.83, -2.17), (-0.83, 0.17, 0.34), (-0.83, 0.17, 0.34)] * [2, 3, 1] =
[(6.34, -1.16, -3.03), (-1.16, 0.22, 0.45), (-1.16, 0.22, 0.45)]

The transition matrix from B to B' can be found by first finding the inverse of matrix B, then multiplying it by matrix B'. The transition matrix from B' to B can be found by first finding the inverse of matrix B', then multiplying it by matrix B. To verify that the two transition matrices are inverses of each other, we can multiply them together and see if the result is the identity matrix. Finally, to find the coordinate matrix [x]B, given the coordinate matrix [x]B', we can multiply the transition matrix from B' to B by the coordinate matrix [x]B'.

(a) Find the transition matrix from B to B'.
P−1 = B^(-1) * B' =
[(-0.04, 0.04, 0.08), (0.08, -0.08, -0.17), (0.08, -0.08, -0.17)] * [(0, −1, 8), (1, 1, −8), (2, 2, −12)] =
[(0.04, -0.12, 0.68), (-0.09, 0.25, -1.43), (-0.09, 0.25, -1.43)]

(b) Find the transition matrix from B' to B.
P = B'^(-1) * B =
[(-0.33, 0.17, 0.08), (0.08, -0.08, -0.17), (0.08, -0.08, -0.17)] * [(14, −2, −6), (−6, 1, 3), (12, −2, −5)] =
[(4.67, -0.83, -2.17), (-0.83, 0.17, 0.34), (-0.83, 0.17, 0.34)]

(c) Verify that the two transition matrices are inverses of each other.
PP−1 = [(4.67, -0.83, -2.17), (-0.83, 0.17, 0.34), (-0.83, 0.17, 0.34)] * [(0.04, -0.12, 0.68), (-0.09, 0.25, -1.43), (-0.09, 0.25, -1.43)] =
[(1, 0, 0), (0, 1, 0), (0, 0, 1)]

(d) Find the coordinate matrix [x]B , given the coordinate matrix [x]B'.
[x]B = P * [x]B' = [(4.67, -0.83, -2.17), (-0.83, 0.17, 0.34), (-0.83, 0.17, 0.34)] * [2, 3, 1] =
[(6.34, -1.16, -3.03), (-1.16, 0.22, 0.45), (-1.16, 0.22, 0.45)]

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Answer:

[tex]\huge\boxed{\sf \angle A = 43\°}[/tex]

[tex]\huge\boxed{\sf \angle B = 59\°}[/tex]

Step-by-step explanation:

In ΔDBC,

98 + ∠B + 31 = 180°

∠B + 121 = 180

∠B = 180 - 121

So,

∠C (Exterior Angle) = ∠A + ∠B

102 = ∠A + 59

102 - 59 = ∠A

43 = ∠A

[tex]\rule[225]{225}{2}[/tex]

The given function 3(t^4)-19 is thus written using the function rule in notations as: f(t) = 3(t^4)-19 for the input-output machine.

Although equations are frequently used to represent functions, a sequence of equations all beginning with y = may make it difficult to distinguish between different functions.

A function called f is defined by the notation f(x).

The given function 3(t^4)-19 is thus written using the function rule in notations as:

f(t) = 3(t^4)-19.

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The equation that represent the function is h(x) = 30x + 50 and the equation is linear.

Ann is coordinating a fashion show and needs to hire a makeup artist. The makeup artist charges 50 dollars booking fee plus an additional 30 dollars per hour .

Let's use function to represent the total cost to hire the artist for x hours.

Therefore,

h(x) = 50 + 30x

Hence, the function is linear.

Therefore, the equation that represent the situation is h(x) = 50 + 30x

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Answer:

n=10

Step-by-step explanation:

(photo linked = step-by-step)

There is no real value of n that satisfies the equation c(n,6) = c(n,4). In other words, there is no real number n such that it satisfy the given combinations.

To find the value of n that satisfies the equation c(n,6) = c(n,4), we can use the formula for combinations.

The formula for combinations, often denoted as C(n, r), calculates the number of ways to choose r items from a set of n items without regard to the order.

In this case, we have c(n, 6) and c(n, 4). The equation states that the number of ways to choose 6 items from a set of n items is equal to the number of ways to choose 4 items from the same set.

Using the formula for combinations, we can set up the equation as follows:

n! / (6!(n-6)!) = n! / (4!(n-4)!)

To solve this equation for n, we can simplify the equation by canceling out the factorials:

(n(n-1)(n-2)(n-3)(n-4)(n-5)) / (6(5)4(3)2(1)) = (n(n-1)(n-2)(n-3)) / (4(3)2(1))

By canceling out common factors and simplifying, we are left with:

(n-5) = (n-4)(n-3)

Expanding the equation, we have:

[tex]n - 5 = n^2 - 7n + 12[/tex]

Rearranging the equation and combining like terms, we get:

[tex]n^2 - 8n + 17 = 0[/tex]

At this point, we can solve the quadratic equation using factoring, the quadratic formula, or other methods. However, upon inspection, we can see that this quadratic equation does not have real solutions.

Therefore, there is no real value of n that satisfies the equation c(n,6) = c(n,4). In other words, there is no real number n such that it satisfy the given combinations.

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Answer:

x > -3

Step-by-step explanation:

10(3x + 2) > 7(2x − 4)

30x + 20 > 14x - 28

16x + 20 > -28

16x > -48

x > -3

So, the answer is x > -3

Therefore , the solution of the given problem of unitary method comes out to be  6.10 is the standard deviation.

The task can be completed by integrating what was learned and putting this variable technique into practise, which also incorporates all supplemental data from two individuals who used a specific tactic. Or, to put it another way, if the desired expression outcome happens, either the entity stated in the algorithm will also be found, or both essential processes will actually bypass the colour. For forty pens, a refundable charge of Rupees ($1.01) might be required.

Here,

Using the ideas of mean and standard deviation, we can interpret the three grades in terms of achievement.

The three categories' mean (average) is:

=> Mean = (75 + 85 + 90) / 3 = 83.33

This indicates that the achievement as a whole is above average (since the mean of 72, 82, and 88 is 80).

Which is:

=> Variance = [(75 - 83.33)² + (85 - 83.33)² + (90 - 83.33)²] / 3

=>  [67.11 + 1.36 + 44.11] / 3

=>  37.19

The variance's square root equals the standard deviation.

=> SD= √(37.19)

=> 6.10 is the standard deviation.

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Hence, there is a 38.2% likelihood that the group chosen will consist of two men and two women.

A committee's primary responsibility is to aid in the smooth operation of the company. The majority of the time, the main concerns of a committee are information sharing and assisting the leadership in making decisions by providing them with the necessary knowledge.

We may employ the formula of combinations to determine the likelihood of choosing a group comprising two men plus two women from of the committee. C(7,2) = 21 ways to choose two males from a group of seven, and C(4,2) = 6 ways to choose two women from a group of four. Hence, there are 21 x 6 = 126 different ways to choose a team of two men and two women.

C(11,4) = 330 different techniques can be used to choose any four members of the committee. Thus, the following is the likelihood of choosing a team of two men and two women:

126 / 330 = 0.382 or 38.2%

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A permutation matrix is a square matrix that has exactly one entry of 1 in each row and each column, and all other entries are 0. When a permutation matrix is multiplied with another matrix, it rearranges the rows of the other matrix.

For matrix A = \begin{bmatrix}-1 & 2 & -2\\ 2 & 3 & 5\\ 1 & 2 & 2\end{bmatrix}

(a) A permutation matrix P that allows for an LU decomposition of PA is P = \begin{bmatrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}

When we multiply P and A, we get PA = \begin{bmatrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}-1 & 2 & -2\\ 2 & 3 & 5\\ 1 & 2 & 2\end{bmatrix} = \begin{bmatrix}2 & 3 & 5\\ -1 & 2 & -2\\ 1 & 2 & 2\end{bmatrix}

(b) An LU decomposition of PA is L = \begin{bmatrix}1 & 0 & 0\\ -0.5 & 1 & 0\\ 0.5 & 0 & 1\end{bmatrix} and U = \begin{bmatrix}2 & 3 & 5\\ 0 & 3.5 & 4.5\\ 0 & 0 & -0.5\end{bmatrix}

So, PA = LU = \begin{bmatrix}1 & 0 & 0\\ -0.5 & 1 & 0\\ 0.5 & 0 & 1\end{bmatrix}\begin{bmatrix}2 & 3 & 5\\ 0 & 3.5 & 4.5\\ 0 & 0 & -0.5\end{bmatrix} = \begin{bmatrix}2 & 3 & 5\\ -1 & 2 & -2\\ 1 & 2 & 2\end{bmatrix}

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The value of x is 4 using the trigonometric function.

The term "hypotenuse" refers to a right-angled triangle's longest side as compared to the base and perpendicular lengths. The right angle, the largest angle of the three angles in a right triangle, lies opposite the hypotenuse side. In essence, only the right triangle possesses the hypotenuse feature, not any other triangles. When we study the right-angled theorem, Pythagoras Theorem, or Pythagorean Theorem, this is better understood. The main use of these ideas is in trigonometry.

The triangle is a right-triangle.

Thus, using the trigonometric functions we have:

Sin (30) = opposite / hypotenuse

sin (30) = 2 / x

1/2 = 2/x

x =2(2)

x = 4

Hence, the value of x is 4 using the trigonometric function.

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Answer:

A. This shape is a rectangular prism — a rectangle with a third dimension of depth. In technical terms, it is "a polyhedron with two congruent and parallel bases." - BYJU's Maths

B. We could find the area of its net, since the surface area of a shape is the sum of the areas of its sides.

The convention centre has a volume of approximately 30,085,616.8 cubic metres as a result.

As opposed to length, width, and height for a rectangle's surface, the basic formula for volumes is length, breadth, and height. It doesn't matter how you refer to the different dimensions; for example, using "depth" instead of "height" wouldn't affect the calculation.

V = base area ×height (1/3)

The pyramid's height is disclosed to be 332 meters. We multiply the base's length and width to determine the base area:

Base area = 475 m × 571 m = 271525 m²

We can now enter the values into to the formula as follows:

V = (1/3) × 271525 m² × 332 m

V = 30085616.7 m³

This is rounding up to the closest tenth, giving us:

V ≈ 30085616.7 ≈ 30085616.8 m³

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The result of blending 50 percent of each animation of the pairs that follow will be a new animation that combines the features of both animations. This is known as "root motion blending" and is used to create smooth transitions between different animations.

The new animation will have a duration that is the average of the two animations being blended (for example, blending IDLE and WALK_1 will result in a 3s duration animation) and will include features from both animations. For example, blending WALK_1 and WALK_2 will result in a walking animation that is at a pace between relaxed and fast. Similarly, blending SPRINT_1 and SPRINT_2 will result in a sprinting animation that includes both a straight sprint and a slight right turn. The root motion of the new animation will be the average of the root motions of the two animations being blended.

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