A line's y-intercept is 9, and its slope is 5. What is its equation in slope-intercept form?
Write your answer using integers, proper fractions, and improper fractions in simplest form.
y=____ x +____

Answer:

Coefficient of x

Write in the slope intercept form

35

Step-by-step explanation:

165 - 105x + 3y = 0  Subtract 165 from both sides and and 105x to both sides

3y = 105x - 165  Divide all the way through by 3

y = 35x - 55

The slope is the coefficient of x when the equation is written in the slope intercept form of the line.

The slope is 35

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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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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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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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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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]

Answer:

Step-by-step explanation:

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:

1. Supplementary

2. Alternate exterior

Step-by-step explanation:

1. ∠1 and ∠2 are complementary because they both lie on the same straight line. If you had the values of these angles, they would add up to 180°

2. ∠1 and ∠8 are alternate exterior angles because they are on opposite sides of the transversal (the diagonal line from the bottom left to the top right) and they are on the outsides of the parallel lines. (∠1 is above the top horizontal line and ∠8 is below the bottom horizontal line.)

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:

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

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 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 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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Answer: C

Step-by-step explanation:

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

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 vertical intercepts is (4/3,0) (3,0) (1,0) and The horizontial intercepts is (0,72)

The end behavior is as x approaches positive infinity, f(x) approaches negative infinity. As x approaches negative infinity, f(x) approaches negative infinity.

Function is a type of relation, or rule, that maps one input to specific single output.

Given the G(x) = -2(3x + 4)(x - 1)(x – 3)^2 as a polynomial function.

Since all the exponets will equal 0. The constant will just add to the term.

The leading degree is even. Thus Even Degree Polynomials like Quadratics tend to go approach positive infinity vertically as x approaches positive infinity, and as x approaches negative infinity, it approaches positive infinity vertically.

Our leading coefficient is negative which means our quadratic will get reflected across the x axis.

The end behavior is as x approaches positive infinity, f(x) approaches negative infinity. As x approaches negative infinity, f(x) approaches negative infinity.

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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:

The distance AB between the two points is 5.7 units

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

A = (2, 2) and B =(-2, -2)

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

distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Where

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

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

distance = √[(2 + 2)² + (2 + 2)²]

Evaluate

distance = 5.7

Hence, the distance is 5.7 units

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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

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 inverse function of f(x) = -3x + 3 in slope-intercept form is g(y) = (-1/3)y + 1.

What is the inverse function?

The inverse function of a function f(x) is a function g(x) such that g(f(x)) = x for all x in the domain of f.

The given function is f(x) = -3x + 3. To find the inverse function, we can solve for x in terms of y:

y = -3x + 3

0 = -3x + y + 3

-y = -3x + 3

x = (-y + 3) / -3

The inverse function is then g(y) = (-y + 3) / -3. Rewriting this in slope-intercept form gives:

g(y) = (-1/3)y + 1

Hence, the inverse function of f(x) = -3x + 3 in slope-intercept form is g(y) = (-1/3)y + 1.

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Answer:

  HAEC

Step-by-step explanation:

You want the code letters associated with lines 1–4 on the graph.

Has a y-intercept of -4, and a rise/run if 1/4. The equation is ...

  y = 1/4x -4  (H)

Has a y-intercept of -4, and a rise/run of 4/1 = 4. The equation is ...

  y = 4x -4  (A)

Has a y-intercept of 3, and a rise/run of -2/1 = -2. The equation is ...

  y = -2x +3  (E)

Has a y-intercept of +1 and a rise/run of 2/1 = 2. The equation is ...

   y = 2x +1  (C)

The code is HAEC.

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:

-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,

For given data,

a) A scatter diagram is as shown below.

b) the coefficient of correlation is, r = 0.63

c) the linear regression equation (least-squares equation of the line) to predict weekly sales from advertising expenditures is : y = 3.2208x + 343.71, where x is the Advertising Costs and y be the Sales in dollars

d) when advertising costs are $32, the weekly sales would be 446.78 dollars.

In this question, we have been given that a retail merchant made a study to determine the relation between weekly advertising expenditures and sales.

The following data were recorded:

Advertising Costs, $ 40 20 25 20 30 50 40 20 50 40 25 50

Sales, $ 385 400 395 365 475 440 490 420 560 525 480 510

a) a scatter plot for the above data is as shown below.

b) the coefficient of correlation is:

r = √0.403

r = 0.63

c) the linear regression equation is:

y = 3.2208x + 343.71

where x is the weekly Advertising Costs and y be the Sales in dollars

d) the weekly sales when advertising costs are $32:

y = 3.2208(32) + 343.71

y = 446.78

So, when advertising costs are $32, the weekly sales would be 446.78 dollars.

Therefore, for given data:

a) A scatter diagram is as shown below.

b) the coefficient of correlation = 0.63

c) the linear regression equation: y = 3.2208x + 343.71

d) when advertising costs are $32, the weekly sales would be 446.78 dollars.

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