claire boarded the airplane in richmond, va, and flew 414 miles directly to Charleston, sc. The total flight time was 3/4 hours. How fast did Claire's airplane fly, in mile per hour

Cost to fill the container = $ 462

The area that any three-dimensional solid occupies is known as its volume. These solids can take the form of a cube, cuboid, cone, cylinder, or sphere. Various forms have various volumes.

The formula for the volume of a rectangular container is given by  

Here we have,

Dimensions of a container are 3yd × 5yd × 7 1/3 yd  

Here 7 1/3 = 22/3 yd

From the given formula,  

Area of the container =  3yd × 5yd × 22/3 yd  

= [tex](3 \times 5 \times \frac{22}{3} )yd^{3}[/tex]

= [tex]( 5 \times 22 )yd^{3}[/tex]      

= 110 yd³  

Given the cost per 1 yd³ = $ 4.20

=> cost of 110 yd³ = 110 × $ 4.20 = $ 462

Therefore,

Cost to fill the container = $ 462

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The resultant answer in roster form is B = {-3, -2, -1, 0, 1, 2, 3, 4, 5}.

In the set-builder form, a short, statement, or formula is written inside a pair of curly braces, as opposed to the roster form, where the listed items are enclosed in a pair of curly braces and separated by commas.

Roster or tabular form: In roster form, all of the components of a set are listed, with commas used to divide them and braces used to enclose them.

For instance, Z = the set of all integers = {…,−3,−2,−1,0,1,2,3,…}.

So, we have:

B = {x:x is an integer and -4 < x < 6}

B has numbers from -4 and 6.

Now, write B in roster form as:

B = {-3, -2, -1, 0, 1, 2, 3, 4, 5}

Therefore, the resultant answer in roster form is B = {-3, -2, -1, 0, 1, 2, 3, 4, 5}.

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Complete question:

Write the following sets in roster form:

B = {x:x is an integer and -4 < x < 6}

For the given hexagonal pyramid, the base of the pyramid is a Hexagon, The height of the pyramid is 13 units, and the vertex of the pyramid is 7.

What is a hexagonal pyramid?

A hexagonal pyramid features isosceles triangles as the faces that join the pyramid together at the top and a hexagonal-shaped base.

Given, for a hexagonal pyramid

a) The base of the pyramid is a Hexagon.

b) The height of the pyramid is 13 units.

c) The vertex of the pyramid is 7.

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The two different if-then statements implied by Theorem 2-13 are:

What is the Parallel Lines & Perpendicular Lines?

Parallel Lines :Two or more lines that lie in the same plane and never intersect or meet each other are known as parallel lines,

Perpendicular Lines are formed when two lines meet each other at the right angle or 90 degrees.

Given: The theorem 2-13 is given.

We have to find what are two different if-then statements implied by Theorem 2-13.

The two different if-then statements implied by Theorem 2-13 are:

Hence, The two different if-then statements implied by Theorem 2-13 are:

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The remainder you got when 3x³ - 5x² - 23x + 24 is divided by x - 3 is -9.

Polynomials are expressions in algebra which consist of both variables and coefficients. Sometimes, variables are also known as indeterminates. Polynomials are classified as monomials, binomials, and trinomials based on the degree of the variables in the expression.

Variables in the monomials, binomials and trinomials have the highest degree equals 1, 2 and 3 respectively.

By doing the long division method, we will bet the quotient as 3x² + 4x - 11 and the remainder equals -9.

Let's check this using division algorithm.

Dividend = 3x³ - 5x² - 23x + 24

Divisor = x - 3

Quotient = 3x² + 4x - 11

Remainder = -9

By division algorithm,

Dividend = (Divisor × Quotient) + Remainder

3x³ - 5x² - 23x + 24  = [(x - 3) (3x² + 4x - 11)] + -9

                                 = [3x³ + 4x² - 11x - 9x² - 12x + 33] + -9

                                 = 3x³ + 4x² - 9x² - 11x - 12x + 33 - 9

                                 = 3x³ - 5x² - 23x + 24

Hence -9 is the remainder of this division process.

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The shortest distance between the two points is √[(c + 5)² + (6 - d)²]

From the question, we have the following points that can be used in our computation:

G = (c, 6) and H = (-5, d)

The distance between the two points can be calculated using the following distance equation

distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Where

(x, y) = (c, 6) and (-5, d)

Substitute (x, y) = (c, 6) and (-5, d) in distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

distance = √[(c + 5)² + (6 - d)²]

Hence, the distance expression is √[(c + 5)² + (6 - d)²]

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5.5km/hr is the speed at which he walks.

Two or more expressions with an Equal sign is called as Equation.

Given

w = walking speed

t = time walking = 2

Given, A man walks for 2 hours at a certain speed which is 2w

He then cycles at three times that seed for 4 hours.

We can form a equation by given data

2w + 3×w×4 = 77

2w+12w=77

14w = 77

Divide both sides by 14

w = 5.5 km/hr

Hence, 5.5km/hr is the speed at which he walks.

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Answer:

D, I've taken the test already!

Step-by-step explanation:

there

The plane is perpendicular to the axis but does not go through the vertex. The plane intersects the double cone at only the vertex is ellipse.

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Answer:

$0.50 per ounce

Step-by-step explanation:

To find the cost per ounce of the chocolate, you will need to divide the price of the box by the weight of the box in ounces. Since there are 16 ounces in a pound, the weight of the box in ounces is 1 * 16 = <<1*16=16>>16 ounces.

To find the cost per ounce, divide the price of the box by the weight of the box in ounces: $8.00 / 16 ounces = $0.50 per ounce.

Therefore, the cost per ounce of the chocolate is $0.50.

Answer: C

Step-by-step explanation:

The problem solving plan is a method to make solving a word problem easier.

The equation of the line that is perpendicular to y = 3x + 1, in​ slope-intercept, is: y = -1/3*x - 2.

If we are given a line that passes a point (12, -6) and is perpendicular to y = 3x + 1, we can find the equation of the line in slope-intercept form following the steps below.

First, find the slope of y = 3x + 1. The slope is 3. Since both lines are perpendicular to each other, therefore, the slope of the line that passes through (12, -6) would be the negative reciprocal of 3, which is m = -1/3.

Substitute m = -1/3 and (a, b) = (12, -6) into y - b = m(x - a):

y - (-6)) = -1/3(x - 12)

y + 6 = -1/3(x - 12)

Rewrite in slope-intercept form:

y + 6 = -1/3*x + 4

y + 6 - 6 = -1/3*x + 4 - 6

y = -1/3*x - 2

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Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

ax² + bx + c = 0

D = b² - 4ac

If D > 0 , then quadratic equation has 2 roots.

If D = 0 , then quadratic equation has 1 roots.

If D > 0 , then quadratic equation has No roots in the set of real numbers.

~~~~~~~~~~~~~~~~~~~~

2y² + 4y = 3

2y² + 4y - 3 = 0

a = 2 , b = 4 , c = - 3

D = 4² - 4(2)( - 3) = 40 > 0 ( 2 solutions )

[tex]y_{12}[/tex] = ( - 4 ± 2√10 ) ÷ 4

[tex]y_{1}[/tex] = [tex]\frac{-2+\sqrt{10} }{2}[/tex]

[tex]y_{2}[/tex] = [tex]\frac{-2-\sqrt{10} }{2}[/tex]

Using the laws of sines and cosines, answers to the questions are as follows,

1. A=52°, b=120, c = 160,  

SAS property,  a=128

2. a=13.7, A=25.43°, B=78°,

AAS property, b=31.21

3. A=38°, B=63°, c=15,

ASA property, b=14

4. a=1.5, b=2.3, c=1.9,

SSS property, B=84

5. b=795.1, c=775.6, B=51.85°,

SAS property, C=50

6. b=40, c=45, A=51°,

SAS property, a=37

8. a=20, b=12, c=28

SAS property, C=120

9. a=125, A=25°, b=150

SAS property, 2 solutions are there, B1=149, B2=30.4

10. b=15.2, A=12.5°, C=57.5°

SAS property, c=14

What are the laws of sine and cosine?

We can determine a triangle's one side's length or one of its angles' measurements using the laws of sine and cosine.
In most cases, the unknown sides or angles of an oblique triangle are calculated using the law of sines formula.
The equation that connects the lengths of the triangle's sides and the cosines of its angles is known as the law of cosine or cosine rule.

1. Given A=52°, b=120, c = 160.

If we construct the triangle, we see that it is satisfying the SAS property

By using the law of cosine we get,

[tex]a^{2}=b^{2} +c^{2}-2.b.c.\cos A\\a=\sqrt{b^{2}+c^{2}-2.b.c.\cos A}\\a=\sqrt{120^{2}+160^{2}-2.120.160.\cos 52}\\a=127.9[/tex]

2. Given, a=13.7, A=25.43°, B=78°

From angle A, angle B, and side a, we calculate side b, by using the Law of Sines,

[tex]\frac{b}{a}=\frac{\sin B}{\sin A}\\b=a.\frac{\sin B}{\sin A}\\b=13.7. \frac{\sin 78}{\sin 25.43}\\b=31.21[/tex]

3. A=38°, B=63°, c=15

From angle A and angle B, we calculate angle C,

[tex]A+B+C=180\\C=180-A-B\\C=180-38-63\\C=79[/tex]

Next, From angle A, angle C, and side c, we calculate side a, by using the Law of Sines

[tex]\frac{a}{c}=\frac{\sin A}{\sin C}\\a=c.\frac{\sin A}{\sin C}\\a=15. \frac{\sin 38}{\sin 79}\\a=9.41[/tex]

Calculation of the third side b of the triangle using a Law of Cosines,

[tex]b^{2}=a^{2} +c^{2}-2.a.c.\cos B\\b=\sqrt{a^{2}+c^{2}-2.a.c.\cos B}\\b=\sqrt{9.1^{2}+15^{2}-2.9.41.15.\cos 63}\\b=13.68[/tex]

4. a=1.5, b=2.3, c=1.9

Calculation of the inner angles of the triangle using a Law of Cosines

[tex]b^{2}=a^{2}+c^{2}-2.a.c.{\cos B}\\B=arc {\cos} \frac{a^{2}+c^{2}-b^{2} }{2.a.c} \\ B=arc {\cos} \frac{1.5^{2}+1.9^{2}-2.3^{2} }{2 . 1.5.1.9}\\B=84.15[/tex]

5. b=795.1, c=775.6, B=51.85°

From angle B, side c, and side b, we calculate side a. by using the Law of Cosines and quadratic equation:

[tex]b^2 = c^2 + a^2 - 2.c. a. {\cos B} \\ 795.1^2 = 775.6^2+a^2-2. 775.6. a . \cos 51\ \\ a > 0 \\ a = 989.175[/tex]

Calculation of angle C of the triangle using a Law of Cosines

[tex]c^{2}=a^{2}+b^{2}-2.a.b.{\cos C}\\C=arc {\cos} \frac{a^{2}+b^{2}-c^{2} }{2.a.b} \\ C=arc {\cos} \frac{989.175^{2}+795.1^{2}-775.6^{2} }{2 . 989.795}\\C=50.5[/tex]

6. b=40, c=45, A=51°

Calculation of the third side a of the triangle using a Law of Cosines

[tex]a^{2}=b^{2} +c^{2}-2.b.c.\cos A\\a=\sqrt{b^{2}+c^{2}-2.b.c.\cos A}\\a=\sqrt{40^{2}+45^{2}-2.0.45.\cos 51}\\a=36.87[/tex]

8. a=20, b=12, c=28

Calculation of angle C of the triangle using a Law of Cosines

[tex]c^{2}=a^{2}+b^{2}-2.a.b.{\cos C}\\C=arc {\cos} \frac{a^{2}+b^{2}-c^{2} }{2.a.b} \\ C=arc {\cos} \frac{20^{2}+12^{2}-28^{2} }{2 . 20.12}\\C=120[/tex]

9.  a=125, A=25°, b=150

2 solutions are possible for this,

solution for B1:

From the angle A, side b, and side a, we calculate side c. by using the Law of Cosines and quadratic equation:

[tex]a^2 = b^2 + c^2 - 2b c \cos A \ \\ 125^2 = 150^2 + c^2 - 2 \cdot \ 150 \cdot \ c \cdot \ \cos 25\degree \ \ \\ c > 0 \ \\ c = 28.21[/tex]

Calculation of angle B of the triangle using a Law of Cosines

[tex]b^{2}=a^{2}+c^{2}-2.a.c.{\cos B}\\B=arc {\cos} \frac{a^{2}+c^{2}-b^{2} }{2.a.c} \\ B=arc {\cos} \frac{125^{2}+28.21^{2}-150^{2} }{2 . 125.28}\\B=149[/tex]

solution for B2:
From the angle A, side b, and side a, we calculate side c. by using the Law of Cosines and quadratic equation:

[tex]a^2 = b^2 + c^2 - 2b c \cos A \ \\ 125^2 = 150^2 + c^2 - 2 \cdot \ 150 \cdot \ c \cdot \ \cos 25\degree \ \ \\ c > 0 \ \\ c = 243.679[/tex]

Calculation of angle B of the triangle using a Law of Cosines

[tex]b^{2}=a^{2}+c^{2}-2.a.c.{\cos B}\\B=arc {\cos} \frac{a^{2}+c^{2}-b^{2} }{2.a.c} \\ B=arc {\cos} \frac{125^{2}+243^{2}-150^{2} }{2 . 125.243}\\B=30.28[/tex]

10.  b=15.2, A=12.5°, C=57.5°

From angle A and angle C, we calculate angle B:

[tex]A+B+C=180\\B=180-A-C\\B=180-12.5-57.5\\B=110[/tex]

From the angle A, angle B, and side b, we calculate side a, by using the Law of Sines.

[tex]\ \\ \dfrac{ a }{ b } = \dfrac{ \sin A }{ \sin B } \\ a = b \cdot \ \dfrac{ \sin A }{ \sin B } \\ a = 15.2 \cdot \ \dfrac{ \sin 12.30 }{ \sin 110\degree } = 3.5[/tex]

Calculation of the third side c of the triangle using a Law of Cosines

[tex]c^{2}=a^{2} +b^{2}-2.a.b.\cos C\\c=\sqrt{a^{2}+b^{2}-2.a.b.\cos C}\\c=\sqrt{15.2^{2}+3.5^{2}-2.15.4.\cos 57.30}\\c=13.64[/tex]

Therefore, we have found the solutions of all of the above bits using the law of sine and cosine.

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Answer:  the cost is $35.51, That must mean it would be driven 33.0 miles.

Step-by-step explanation: The given function c(r) represents the cost of a one-day truck rental when the truck is driven x miles. This means that the cost in dollars is a linear function of the number of miles driven.

To answer the first part of the question, we can plug in x = 80 into the function to find the truck rental cost when we drive 80 miles:

$$c(r) = 0.47x + 20 = 0.47(80) + 20 = \boxed{35.6}$$

To answer the second part of the question, we can solve the equation $c(r) = 0.47x + 20 = 35.51$ for x to find the number of miles driven when the cost is $35.51:

$$35.51 = 0.47x + 20 \Rightarrow 15.51 = 0.47x \Rightarrow x = \boxed{33.0}$$

Therefore, when the cost is $35.51, we must have driven 33.0 miles.

The required slope of the side NO is given as 2/3 and the Slope of the LO is given as -3/2.

Given that,
A figure shown of a quadrilateral in order to prove the quadrilateral as a rectangle, the slope of the sides LO and NO is to be determined.

The rectangle is 4 sided geometric shape whose opposites are equal in length and all angles are about 90°.

here,

Because of the parallel sides,
The slope of the side NO is  = Slope of the side LM
                                            = [-1 + 5] / [4 + 2] = 2/3

Because of the perpendicular sides,
The slope of the side LO = - 1 / slope of the side LM
                                = -1 / 2/3 = -3/2

Thus, the required slope of the side NO is given as 2/3 and the Slope of the LO is given as -3/2.

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Estimated volume of the tree as per given height and circumference using trapezoidal rule is equal to 30907.5 cubic inches.

As given in the question,

Given height 'x' and circumference 'y=f(x)',

Height (inches)  'x'            :       0    15     30    45      60      75      90

Circumference (inches) 'y=f(x)':      31   28    21    17      12      8        2

Trapezoidal rule :

Δx = (b - a ) n

Here b = 90

a = 0

n =6

Δx = ( 90 - 0)/6

    = 15

Substitute the value to get the volume using trapezoidal rule:

T₆=(Δx/2)[f(x₀)²+ 2{f(x₁)²+ f(x₂)²+f(x₃)² + f(x₄)²+f(x₅)²}+ f(x₅)²]

   = (15/2)[ 31² + 2 (28² + 21² + 17² + 8²) + 2² ]

   = ( 15/2) [961 + 2{ 784+ 441 + 289 + 64} + 4]

   = 15 × 2060.5

   =   30907.5 cubic inches

Therefore, the volume of the tree as per given given table using trapezoidal rule is equal to 30907.5 cubic inches.

The above question is incomplete , the complete question is:

The circumference of a tree at different heights above the ground is given in the table below. Assume that all horizontal cross-sections of the tree are circles. Estimate the volume of the tree using the trapezoid rule. There needs to be six subdivisions in the trapezoid rule.

Height (inches) :                  0    15     30    45      60      75      90

Circumference (inches):      31   28    21    17      12      8        2

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a) The Domain is [-∞, 5]

And the range is [-∞, 3]

b) The area covered by a relation's inverse is the same as its domain. In other words, the relationship's x-values are its inverse's y-values.

c) The Domain is [-∞, 5]

And the range is [-∞, 3]

d) The graph of R is shown below:

The set of inputs that a function will accept is referred to as the domain of the function in mathematics. It is important to note that in contemporary mathematical terminology, a function's domain is a component of its definition rather than a quality.

The function f can be plotted in the Cartesian coordinate system in the special case where X and Y are both subsets of R. In this example, the graph's x-axis shows the domain as the projection.

Given,

R = {(x, y): y=x²-4 and y ≤5]

For x=0, y=-4

For x=1, y=-3

For x=2, y=0

For x=3, y=5

Therefore, the Domain is [-∞, 5]

And the range is [-∞, 3]

The area covered by a relation's inverse is the same as its domain. In other words, the relationship's x-values are its inverse's y-values.

The graph of R is shown below:

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On solving the provided question, by help of the Linear Equation we got that TV = 38 units.

Any equation with a degree of 1 or above is considered linear. This shows that the exponent of the linear equation's variable is bigger than 1. A linear equation will always have a straight line as its graph.

RST and RSV, two equal right triangles, are shown in the illustration.

Notice that:

TS = VS

We know that,

TS = 6x - 3

VS = 39

6x - 3 =  39

6x = 42

[tex]x = \frac{42}{6}[/tex]

x = 7

Now, you can identify in the figure that:

RV = 2x + 5

The length of segment RV may be determined by substituting the above-calculated value of "x" into the equation and evaluating:

RV = 2(7)+ 5

RV = 19 units

Now,

TV = 19 + 19

TV = 38 units

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The solution to the system of equations 3x+y=24 and -x-y=-10 is x = 7, y = 3

The system of equations is given as

3x+y=24

-x-y=-10

Make x the subject in the second equation

So, we have the following representation

x = 10 - y

Substitute x = 10 - y in the equation 3x+y=24

3(10 - y) +y=24

Expand the bracket

30 - 3y + y = 24

Evaluate the like terms

-2y = -6

Divide by -2

y = 3

Recall that

x = 10 - y

So, we have

x = 10 - 3

x = 7

Hence, the solution is x = 7, y = 3

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Step-by-step explanation:

If M is the midpoint of AB then,

AM = BB so we can write the following equation:

4x + 13 = 3x + 17

4x - 3x = 17 - 13

x = 4

Now on to the length of BM, we can replace x with 4 to find it.

3*4 + 17 = 29

Answer:

Angel 5 is 73

Angel 1 is 37

Angel 2 is 42

Angel 3 is 132

Angel 4 is 73

Angel 6 is 30

Step-by-step explanation:

Answer:

Step-by-step explanation:

HORIZONTAL ASYMPTOTE

lim x-> oo  -5/(2x-1)

-5/(oo-1)

-5/oo

0

y = 0

VERTICAL ASYMPTOTE

2x - 1 = 0

2x = 1

2/2 x = 1/2

x = 1/2

y INTERCEPT

-5/(0-1)

-5/-1

5

(5,0)

X INTERCEPT

-5 = 0

Impossible

no x intercept

Answer:

-6v-1+v

collect the like terms

-6v + v - 1

you have

-5v-1

Answer:

answer is 5v-1

Step-by-step explanation:

add the common varribles which are V,

Answer:

Step-by-step explanation:

You want the number of tickets of each kind sold if 169 tickets were sold for $1795, and lower box tickets were $12.50 while upper box tickets were $10.

Let x represent the number of lower box tickets sold. Then 169 -x is the number of upper box tickets sold. The total revenue is ...

  12.50(x) + 10.00(169 -x) = 1795.00

Simplifying, we get

  2.50x +1690 = 1795

  2.5x = 105 . . . . . . . . . . . subtract 1690

  x = 42 . . . . . . . . . . . . divide by 2.5; the number of lower box ticket sold

  169 -42 = 127 . . . . . . the number of upper box tickets sold

42 lower box and 127 upper box tickets were purchased.