Which of the options is a solution for the following system of inequalities?
2x+y >3
x-y >1
O (5,-2)
O (0,-5)
O (0,0)
O (0,5)

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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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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Answer:

91

Step-by-step explanation:

[tex]\frac{89+82+69+79+70+x}{6}[/tex] = 80  Combine like terms

[tex]\frac{389+ x}{6}[/tex] = 80  Multiple by sides by 6

380 + x = 480  Subtract 389 from both sides.

x = 91

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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The system of equations is y = 4x - 3 and y = -1/3x + 4 and the solution is (21/13, 45/13)

From the question, we have the table of values as the parameters that can be used in our computation:

From the table, we have

Table 1

(x, y) = (0, -3) and (4, 13)

A linear equation is represented as

y = mx + c

Where

Slope = m

c = y when x = 0

This means that

c = -3

So, we have

y = mx - 3

The point (4, 13) implies that

13 = m * 4 - 3

So, we have

4m = 16

m = 4

So, the equation is

y = 4x - 3

Table 2

(x, y) = (0, 4) and (4, 6)

A linear equation is represented as

y = mx + c

This means that

c = 4

So, we have

y = mx + 4

The point (4, 6) implies that

6 = m * 4 + 4

So, we have

4m = -2

m = -1/3

So, the equation is

y = -1/3x + 4

Substitute y = -1/3x + 4 in y = 4x - 3

-1/3x + 4 = 4x - 3

So, we have

-x + 12 = 12x - 9

Evaluate the like terms

13x = 21

Divide

x = 21/13

y = -1/3x + 4 implies that

y = -1/3 x 21/13 + 4

So, we have

y = -7/13 + 4

Evaluate

y = 45/13

So, the solution is (21/13, 45/13)

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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

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}

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 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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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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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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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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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:

Answer: C

Step-by-step explanation:

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

Application acceptance requires a minimum score of 392, which is required.

Make "x" the required minimum score.

Given:

The average score is 500.

Standard deviation () is equal to 100

Admission percentage, P > 86 percent, or 0.86

Thus, the area under the normal distribution curve to the right of the z-score, which is 86%, is provided.

The portion of the score left over is displayed in the z-score table. As a result, we will calculate the z-score value for area as 100 - 86 = 14% or 0.14.

The z-score is therefore equal to -1.08 for value of 0.1401.

P(x>x0) = P(z > -1.08)

= -1.08 = x0 - 500 /100

x0 - 500 = -1.08 × 100

x0 = -1.08 + 500 = 392.

As a result, 392 is the minimal score needed for admission.

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Answer: 475 minutes.

Step-by-step explanation: To compare the cost of these two plans, you need to find how many minutes of calls are needed for the cost of each plan to be the same. Let's call this number of minutes x.

For plan A, the total cost will be $19 plus $0.11 for each minute of calls, or a total of 19 + 0.11x dollars.

For plan B, the total cost will be $0.15 for each minute of calls, or a total of 0.15x dollars.

Since the cost of the two plans is equal, we can set these expressions equal to each other and solve for x:

19 + 0.11x = 0.15x

Subtracting 0.11x from both sides, we get:

19 = 0.04x

Dividing both sides by 0.04, we get:

475 = x

Therefore, the number of minutes of calls needed for the cost of the two plans to be the same is 475 minutes.

Based on the fact that Hannah's investment will triple in 10 years and 10 months when compounding occurs every six months.

The process through which interest is added to the existing principal sum and the interest that has previously been paid is known as compounding.

As a result, compounding, sometimes known as the "magic of compounding," can be thought of as interest on interest, which has the effect of making returns on interest larger over time.

The amount of time required to triple the investment is as follows:

$2500 was invested.

3.4% interest rate.

Compounding: Every 6 months (Semiannually)

Now think about a time frame of two years:

3.4 * 6/12 = 1.7%

The effective interest rate will be 1.7%.

The current value of the 1.7% compounded rate over 65 factors is equivalent to 2.9913.

The value will therefore be $2,500 2.9913 = 7,478.25.

Six years divided by 65 months yields a total of 10.833 years.

10 months make up a year, or 0.833 * 12.

Therefore, based on the fact that Hannah's investment will triple in 10 years and 10 months when compounding occurs every six months.

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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:

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.

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:

D, I've taken the test already!

Step-by-step explanation:

there

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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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.

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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