Which statement is true about this data set?
3,1, 6, 4, 4, 4, 3, 2, 5, 3

The answer choice which describes the function is; Choice B; neither symmetry about the y-axis nor has order 2 rotational symmetry about the origin.

Rotational symmetry, otherwise termed radial symmetry, is the characteristic of a shape when it looks the same after some rotation by a partial turn. On this note, since the function represents a quartic graph, it follows that the function has neither symmetry about the y-axis nor has order 2 rotational symmetry about the origin.

Therefore, the function f(x) = −4x⁴ + 3x³ − 2x² − x + 5 is neither symmetry about the y-axis nor has order 2 rotational symmetry about the origin.

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Answer:

  x = 3

  y = 8

Step-by-step explanation:

A matrix product exists if and only if the number of columns in the left matrix is equal to the number of rows in the right matrix.

For AB to exist, we must have ...

  columns of A = rows of B

  x +5 = y

For BA to exist, we must have ...

  columns of B = rows of A

  11 -y = x

Using the second equation to substitute for x in the first equation, we have ...

  (11 -y) +5 = y

  16 = 2y

  8 = y

Then the value of x is ...

  11 -8 = x = 3

The values of x and y are ...

__

Additional comment

The dimensions are A = 3×8; B = 8×3.

The box plot that represents the data is a box plot titled "Scores of Participants" and labeled "Score" uses a number line from 10 to 35 with primary markings and labels at 10, 15, 20, 25, 30, and 35. The box extends from 13 to 27 on the number line. A line in the box is at 24. The whiskers end at 11 and 31.(second option)

A box plot is used to study the distribution and level of a set of scores. The whiskers represent the minimum and maximum values.

On the box, the first line to the left represents the lower (first) quartile. The next line on the box represents the median. The third line on the box represents the upper (third) quartile. 75% of the scores represents the upper quartile.

The data arranged in ascending order : 11, 13, 23, 24, 24, 27, 31

Median = 24

First quartile = 1/4 x (7 + 1) = 2nd term =13

Third quartile = 3/4 x (7 + 1) = 6th term = 27

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Answer: False

Step-by-step explanation:

If the function has a negative slope, then it will decrease.

Answer:

-3

Step-by-step explanation:

[tex]f\left( x\right) =\sqrt{3x} \ \ \text{and} \ \ g\left( x\right) =x+3[/tex]

it’s clear that :

f is defined if and only if  x ≥ 0

Therefore ,

fºg is defined if and only if x + 3 ≥ 0

x + 3 ≥ 0

⇔ x ≥ -3

Then ,the smallest number that is in the domain of fºg is -3

Answer:

Step-by-step explanation:

eq. of line is 2x-3y=-3

its slope=(-co-efficient of x/co-efficient of y)=-2/-3=2/3

slope of line perpendicular to given line=-3/2

eq. of line through (-3,0) is

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

2y=-3x-9

3x+2y=-9

or

any eq. perpendicular to given line is

3x+2y=c where c is a constant.

it passes through (-3,0)

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

c=-9

reqd. eq. is

3x+2y=-9

Answer:

[tex]4a^{4} -11a^{3} + 6a^{2}[/tex]

Answer:

  (a)  100 meters

Step-by-step explanation:

You want the width at the base of a cooling tower whose walls are modeled by the equation (x/20)² -((y-110)/48)² = 1.

The width at the base will be the difference in the values of x when y=0.

  (x/20)² -((0 -110)/48)² = 1

  (x/20)² = 1 +3025/576

  x = ±20/24√3601 ≈ ±50.01

The width will be 50.01 -(-50.01) ≈ 100 meters.

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The equation between the number of days remaining and number of hours is given as y= -5x+50.

Let y be the number of days remaining and x be the no.of hours.

From the graph the slope of the line is given by m.

m=(y2 - y1)/(x2 - x1)

m=(30 - 40)/(4 - 2)

m= -10/2 = -5.

The above slope is in between points 2 and 4. As the line has a constant rate of growth the slope is the same between any points of the line.

Therefore the equation between the no.of days remaining and no.of hours is given by

y=mx +c.

y=-5x+c

Now to find the value of c. This can be determined by substituting the x and y values at a particular instant. For x=0,y=50. Therefore by these values, we get,

50= -5(0)+c

c=50.

Therefore the relation between the number of days remaining and number of hours is given as y= -5x+50.

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Based on the triangle proportionality theorem, the true proportion is h/j=k/i

In the diagram that we have here, we have TU to be parallel to QR.

We have following

ST is equal to h, TQ is equal to j, SU is equal to k and UR is equal to i.

This can then be used to form the equation

ST/TQ = SU/UR

When we h/j = k/i

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The function that's illustrated in the graph is function D.

First, you look at the left side of the graph. You can tell by the shape that it is an x³ function.

Also, the filled-in circle means inclusive. so we know that when x is less than or equal to a number.

The answer that illustrates the information explained is D.

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Answer:

B.

Step-by-step explanation:

An open circle means to exclude the point.

A closed circle means to include the point.

In this graph we see a part of a parabola with an open circle above. The parabola starts at x = 1, but does not include x = 1, and the x values of the parabola increase.

A parabola has an x squared in the equation.

The solution has to have an equation with an x² as its greatest exponent, and also must include x > 1.

The lower part of the graph is a third degree polynomial, so it must have an x cube term, x³. Also, the greatest x value for the x³ function is x = 1, but x = 1 is included. Then all x values tot he left of x = 1 are also included, so for the cubic polynomial, the domain (x values) must be x ≤ 1.

The answer is B.

The nonreal solutions to the polynomial equation are: (2 + i√11, 2 - i√11).

By critically observing the graph given, we can logically deduce that the real solutions of the polynomial are -1 and 4. This ultimately implies that, the given polynomial expression can be evaluated as follows:

(x + 1)(x - 4) = x² - 3x - 4.

x⁴ - 7x³ + 23x² - 29x - 60 = x² - 3x - 4.

Also, the nonreal solutions can be determined by dividing the polynomial expression by the real solutions as follows:

Nonreal solutions = (x⁴ - 7x³ + 23x² - 29x - 60)/(x² - 3x - 4)

Nonreal solutions = (2 + i√11, 2 - i√11).

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The plotting of the inequalities was done using geogebra online graphing calculator. The graph is attached.

Inequalities are the mathematical expressions in which both sides are not equal. It should be noted that in inequality, unlike in equations, we compare two values.

The equal sign in between is replaced by less than (or less than or equal to), greater than (or greater than or equal to), or not equal to sign. Inequality is an expression used to show the nonequal comparison of numbers and variables.

Given the system of inequalities:

3y > 2x + 12

3y - 2x > 12   (1)

2x + y < -5    (2)

Plotting the inequalities using GeoGebra online graphing calculator. The graph is attached

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Answer: D

Step-by-step explanation:

There are vertical asymptotes at x=7 and x=-3, so the denominator should have factors of (x-7) and (x+3).

This eliminates everything except for D.

The probability of pulling a blue or yellow card, written as a reduced fraction is 7/50

From the question, we have the following probabilities:

P(Pink) = 1/5

P(Blue) = 1/25

P(Yellow) = 1/10

P(Black) = 1/15

The probability of pulling a blue or yellow card, written as a reduced fraction is the calculated as:

P(Blue or yellow card) = P(Blue card) + P(Yellow card)

Substitute the known values in the above equation

P(Blue or yellow card) = 1/25 + 1/10

Express 1/25 as 2/50 and 1/10 as 5/10

P(Blue or yellow card) = 2/50 + 5/50

Take the LCM

P(Blue or yellow card) = (2+ 5)/50

Evaluate the sum

P(Blue or yellow card) = 7/50

Hence, the probability of pulling a blue or yellow card, written as a reduced fraction is 7/50

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Step-by-step explanation:

<1 congruent to <5.

<5 congruent to <7.

<1 congruent to <7.

Answer:

don't know but i need thee points

Step-by-step explanation:

The possible length of the third side will be values that are less than 41 and greater than 5

For three sides to be the sides of a triangle, the sum of any two sides of the triangle must be greater than the third.

Given the two sides of a triangle have lengths 18m and 23m, the sum of the sides is given as:

sum = 18 + 23

Sum = 41

Let the third side be x, such that;

x + 18 >23

23 + x > 18

x >  23 - 18

x > 5

Similarly

x > -5

Hence the possible length of the third side will be values that are less than 41 and greater than 5

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Answer:

x = 73

x = -35

Step-by-step explanation:

Here are the 8 numbers:

34, 31, 15, x, 17, 4, 20, 25

Here are the 8 numbers in ascending order with x as the greatest number:

4, 15, 17, 20, 25, 31, 34, x

x - 4 = 69

x = 73

Here are the 8 numbers in ascending order with x as the smallest number:

x, 4, 15, 17, 20, 25, 31, 34

34 - x = 69

-x = 69 - 34

-x = -35

x = -35

Answer:

x = 73

x = -35

The expression which is equivalent to the given expression in the task content is; 1/x²⁴.

It follows from the task content that the given expression is; ((x)^-6/x²)³.

Hence, it follows from the laws of indices that the expression can be evaluated as follows;

= (x^-8)³

= x^-²⁴

= (1/x²⁴).

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The percentage of vehicle mileage expectancies that are more than 188,600 miles is 93%

The given parameters are:

Mean = 169200

Standard deviation = 19400

x = 188600

Calculate the z value using:

[tex]z = \frac{x - \mu}{\sigma}[/tex]

So, we have:

[tex]z = \frac{188600 - 169200}{19400}[/tex]

Evaluate

z = 1

Using the Empirical Rule, we have:

P(x > 188600) = P(z > 1.5)

From z table of probability, we have:

P(x > 188600) = 0.93319

Express as percentage

P(x > 188600) = 93.319%

Approximate

P(x > 188600) = 93%

Hence, the percentage of vehicle mileage expectancies that are more than 188,600 miles is 93%

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The system of equations that could be used to determine the number of small cars rented and the number of large cars rented is;

x + y = 10

x + y = 105x + 7y = 66

Let

Number of small cars rented = x

Number of large cars rented = y

x + y = 10

x + y = 105x + 7y = 66

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Answer:

2 hours

Step-by-step explanation:

If Thomas can paint 1/3 = 2/6 of the room in a hour and Brian can paint 1/6 of the room in an hour. Which means in an hour they can collectively paint 3/6 = 1/2 of the room. Now you need to find how long it would take then to paint 2/2 of the room. Knowing that 1/2+1/2 = 2/2, it would take them 2 hours to paint the room together

The point which would lie on the graph of f(x) = log2x is; (1, 0.301).

Since it follows from the task content that the given point (–1, 0.5) lies on the graph of f –1(x) = 2x.

The subsequent graph of f(x) = log2x would result from taking logarithms in which case, we have the graph passing through points determined as follows;

f(x) = log2^x

f(x) = x log2

Hence, when x = 1;

f(1) = 1log 2

Hence, we have point; (1, 0.301).

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The equation whose solution is the given quantity is 90,000,000 = 2,000,000w + 15,000w where w = $44.67

B = xz + yw

Where,

B = xz + yw

90,000,000 = (20 × 100,000w) + (15,000 × w)

90,000,000 = 2,000,000w + 15,000w

90,000,000 = 2,015,000w

w = 90,000,000 / 2,015,000

= 44.6650124069478

Approximately,

w = $44.67

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Using equivalent angles, the solutions to the equation are given as follows:

[tex]\theta = 255^\circ, \theta = 285^\circ[/tex]

Each angle on the second, third and fourth quadrants will have an equivalent on the first quadrant.

In this problem, the equation is:

[tex]\sin{\theta} = -0.966[/tex]

Applying the inverse relation:

[tex]\theta = \arcsin{-0.966} = 255^\circ[/tex]

Which is on the third quadrant. The equivalent angle on the fourth quadrant is given as follows:

[tex]270^\circ + (270^\circ - 255^\circ) = 285^\circ[/tex]

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The depreciation expense per unit of production for the Textile Company is D. $0.20.

The unit-of-production depreciation method uses the useful life of an asset based on the units of production and the depreciable amount to estimate the depreciation expense per unit.

For each period, the depreciation expense per unit is multiplied by the units of production to determine the period's total depreciation expense.

Cost of equipment = $40,000

Residual value = $6,000

Depreciable amount = $34,000 ($40,000 - $6,000)

Estimated useful life = 170,000 units

Depreciation expense per unit of production = $0.20 ($34,000/170,000)

Thus, the depreciation expense per unit of production for A Textile Company is D. $0.20.

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The mean distance increased by a factor of 2^(²/₃)

The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is;

T² = A³.

Let the orbital period of planet X = T(X).

Let the orbital period of planet Y = T(Y).

Let the mean distance of planet X from the sun = A(X).

Let the mean distance of planet Y = A(Y).

Thus;

A(Y) = 2A(X)

Therefore;

[T(Y)]² = [A(Y)]³ = [2A(X)]^3

However, [T(X)]² = [A(X)]³

Thus;

[T(Y)]² = 2³[T(X)]² * [T(Y)]²/[T(X)]² = 2³T(Y)/T(X) = 2^(³/₂)

That's for orbital period but the mean distance will increase by 2^(²/₃)

Thus we conclude that, the mean distance increased by a factor of 2^(²/₃)

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Answer:

2 Superscript three-halves

Step-by-step explanation:

Edge

The expression x²-10x-21 is wrong since the product of two negative signs is equal to positive.

Quadratic equations are equations that has a leading degree of 2. The standard quadratic equation is given as;

f(x) = ax^2+bx+c

Given the product of the expression below;

f(x) = (x-3)(x-7)

Expand

f(x) = x(x) - 7x - 3x -3(-7)

f(x) = x^2 - 7x - 3x + 21

f(x) = x^2 - 10x + 21

Hence the expression x²-10x-21 is wrong since the product of two negative signs is equal to positive.

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Answer:

ΔABC ≅ ΔEDF

Step-by-step explanation:

Triangle ABC is congruent to triangle EDF because they're the same size.

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

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

[tex] \sf\triangle ABC \cong \triangle DEF[/tex]

For checking congruency, we need to follow a proper order when choosing the triangle. because each triangle name is unique in itself.

So, According to Triangle ABC, we need to find the triangle with first two letters That resembles AB, that is :

and next should be :

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

so, now we got the sense for naming, therefore the answer will be :

[tex] \sf\triangle ABC \cong \triangle DEF[/tex]