Suppose that 4 ≤ f '(x) ≤ 5 for all values of x. what are the minimum and maximum possible values of f(8) − f(3)?

Answer:

36

Step-by-step explanation:

12 times 2 which is 24 then u add 12 cause it's the amount they have together so you get 36

The system of equations that can be used to determine whether the boat comes within 5 miles of the lighthouse is:

To find the system of equations that can be used to determine whether the boat comes within 5 miles of the lighthouse:

The vertex form of a quadratic function is given by: f(x) = a(x - h)^2 + k

Where (h, k) is the vertex of the parabola, a is constant.

For the sailboat we have vertex: (h, k) = (2, -6) and one point: (-7, 8) that is f(-7) = 8.

We will find a using f(-7) = 8:

The quadratic function for the sailboat is given by:

The equation for a circle with a radius of 5 and center (1, 2) is:

Therefore, the system of equations that can be used to determine whether the boat comes within 5 miles of the lighthouse is:

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The correct form of the question is given below:
A lighthouse is located at (1, 2) in a coordinate system measured in miles. a sailboat starts at (–7, 8) and sails in a positive x-direction along a path that can be modeled by a quadratic function with a vertex at (2, –6). which system of equations can be used to determine whether the boat comes within 5 miles of the lighthouse?

Answer:

1st option

Step-by-step explanation:

to find f(g(x)) substitute x = g(x) into f(x) , that is

f(g(x))

= f(4x - 5)

= 2(4x - 5) + 1 ← distribute parenthesis

= 8x - 10 + 1

= 8x - 9

Answer:

Step-by-step explanation:

1. $135

2. a. no

    b. $340

3. a. 33

   b. $135*33= $4455+340= $4795

4. a. $7630 = 135v + 340

     b. 135v + 340 = 7630

          135v = 7290

              v = 54 vacuums        

Step-by-step explanation:

since the 2 arc angles cover the full circle, we know their sum must be 360°.

so,

23x - 5 + 37x + 5 = 360

60x = 360

x = 6

and so the angle at V = 5×6 + 17 = 30+17 = 47°

The multiplication shows that the answers will be:

1. 3744

2. 8244

3. 3630

4. 2616

5. 20772

6. 7820

7. 3916

8. 3521

9. 21285

10. 54162

11. 4548

12. 2091

13. 17128

14. 55692

15. $895

16. 4300 miles

1. 416 × 9 = 3744

2. 1374 × 6 = 8244

3. 726 × 5 = 3630

4. 872 × 3 = 2616

5. 2308 × 9 = 20772

6. 1564 × 5 = 7820

7. 4 × 979 = 3916

8. 503 × 7 = 3521

9. 5 × 4257 = 21285

10. 6018 × 9 = 54162

11. 758 × 6 = 4548

12. 3 × 697 = 2091

13. 2141 × 8 = 17128

14. 7 × 7956 = 55692

15. From the information given, the cost of each ticket is $179. Also, there are 5 people that are flying.

Therefore, the total cost will be:

= $179 × 5

= $895

16. Also, for the second question, since the distance between the two cities is 2150 miles. The exact distance will be:

= 2150 × 2

= 4300 miles.

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The equation of the circle can be shown as, x² + y² = 9², or, x² + y² = 81.

The equation of a circle, with the center at the origin and the radius r units, is given as x² + y² = r².

In the question, we are asked to write the equation for a circle centered at the origin with x-intercepts of (-9, 0) and (9, 0).

We can find the radius of the circle using the distance formula,

D = √((x₂ - x₁)² + (y₂ - y₁)²), where (x₁, y₁) and (x₂, y₂) are the endpoints of a line segment.

Radius is the distance between the center and any point on the circle.

Thus, taking (x₁, y₁) as (0, 0), the origin, for the center, and (x₂, y₂) as (9, 0), for any point on the circle, we get the radius as:

r = √((9 - 0)² + (0 - 0)²),

or, r = √(9² + 0²),

or, r = √9² = 9.

Thus, the radius of the circle is r = 9 units.

Thus, the equation of the circle can be shown as, x² + y² = 9², or, x² + y² = 81.

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Answer:

10

Step-by-step explanation:

f(x) = 3x + 1

f^-1 = x-1/3 , it is known as inverse function.

f(3)= 3*3 +1

=9+1

=10

To find inverse,

x=3x + 1

y=3x+1

Exchanging the position or x and y

x=3y+1

x-1=3y

.•. y=x-1/3

The area of the region is: 63.585 square inches.

If we have a circle of radius R, the area of said circle is:

A = pi*R^2

Particularly, if we have a section of the circle defined by an angle θ, the area of that region is:

A = (θ/2pi)*pi*R^2 = (θ/2)*R^2

In this case we have:

θ = 90° = pi/2

R = 9in

Replacing that we get:

A = (pi/4)*(9in)^2 = (3.14/4)*(9in)^2 = 63.585 in^2

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Answer:

70 inches

Step-by-step explanation:

the width of the original rectangle:

48/2 - 13 = 24 - 13 = 11

the new rectangle:

length: 13

width: 2(11) = 22

perimeter: 2(13+22) = 2(35) = 70

The perimeter of the new rectangle is 70 inches.

The area of Rectangle is length times of width.

The perimeter of a rectangle is given by the formula P = 2(l + w)

where P is the perimeter, l is the length, and w is the width.

The first rectangle has a perimeter of 48 inches and a length of 13 inches.

48 = 2(13 + w)

24 = 13 + w

w = 11

So the first rectangle has a length of 13 inches and a width of 11 inches.

The second rectangle has the same length of 13 inches and twice the width of the first rectangle, which means it has a width of 22 inches.

P = 2(13 + 22)

= 2(35)

= 70

Therefore, the perimeter of the new rectangle is 70 inches.

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The graph of the function has no asymptotes.

The function is given as:

c(t) = 5t - t² + 1

When t = 1, 2, 5, 10;

We have:

c(1) = 5(1) - 1² + 1 = 5

c(2) = 5(2) - 2² + 1 = 7

c(5) = 5(5) - 5² + 1 = 1

c(10) = 5(10) - 10² + 1 = -49

So, the table of values is

t        c(t)

1         5

2        7

5        1

10      -49

See attachment for the graph of the function

The domain

From the question, we understand that t represents time.

So, the domain is 0 ≤ t ≤ 5.2

The range

From the graph, the maximum of the graph is 7.25

So, the range is 0 ≤ f(t) ≤ 7.25

Asymptotes

From the graph, we can see that the graph has no asymptotes.

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The expression which correctly sets up the quadratic formula to solve the equation is (A) [tex]\frac{-(-4)+-\sqrt[]{-4^{2}-4(1)(3) } }{2(1)}[/tex].

What is an expression?

To find which expression correctly sets up the quadratic formula to solve the equation:

Theory of quadratic equation - A quadratic equation is defined as any equation containing one term in which the unknown is squared and no term in which it is raised to a higher power.

An example of a quadratic equation in x is [tex]-4x^{2} +4=9x[/tex].

How to solve any quadratic equation using the Sridharacharya formula?

Let us represent a general quadratic equation in x, [tex]ax^{2} +bx+c=0[/tex] where a, b and c are coefficients of the terms.

According to the Sridharacharya formula, the value of x or the roots of the quadratic equation is -

[tex]x=\frac{-b+-\sqrt{(b)^{2}-4(a)(c) } }{2a}[/tex]

The given equation is [tex]x^{2} -4x+3=0[/tex]

Comparing with the general equation of quadratic equation, we get a = 1, b = -4 , c = 3.

Putting the values of coefficients in the Sridharacharya formula,

[tex]\frac{-(-4)+-\sqrt[]{-4^{2}-4(1)(3) } }{2(1)}[/tex]  which is (A).

Therefore, the expression which correctly sets up the quadratic formula to solve the equation is (A) [tex]\frac{-(-4)+-\sqrt[]{-4^{2}-4(1)(3) } }{2(1)}[/tex].

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The complete question is shown below:

Consider this quadratic equation. x^2+3=4x. Which expression correctly sets up the quadratic formula to solve the equation?

Answer:

E. >

Because 65 is greater than 56.

Answer: 12 cm

Step-by-step explanation:

[tex]\mathtt{Using\ the \ pythagorean \ theorem: \\}[/tex]

[tex]a^2+b^2=c^2, \\\ we\ can\ plug\ in\ values\ for\ a,\ b,\ and\ c[/tex]

[tex]L=a\ and\ M=b\ and\ N=c[/tex]

Therefore,

[tex]L^2+35^2=37^2\\L^2+1225=1369\\L^2=144\\\sqrt{L^2} =\sqrt{144} \\L=\pm12\\Taking\ only\ the\ positive\ answer,\ we\ get:\\\large\boxed{L = 12cm}[/tex]

We have: L² + M² = N²

=> L² = N² - M² = 37² - 35² = 144 = 12²

=> L = 12

ANSWER: B.12

OK done. Thank to me :3

The area of the triangle rounded to the nearest tenth is 33.3 squared inches.

Given the data in the diagram;

First we find the dimension of side b.

From the rule of cosines.

b = √[ a² + c² - 2acCosB ]

We substitute into the formula.

b = √[ 7² + 13² - ( 2 × 7 × 13 × cos( 133° ) ]

b = √[ 49 + 169 - ( 182 × cos( 133° ) ) ]

b = √[ 218 - 182×cos( 133° ) ]

b = √[ 342.1237 ]

b = 18.5

Next, we find angle C.

From rule of cosines.

cosC = [ b² + a² - c² ] / 2ba

cosC = [ 18.5² + 7² - 13² ] / [ 2 × 18.5 × 7 ]

cosC = [ 342.25 + 49 - 169 ] / [ 259 ]

cosC = [ 222.25 ] / [ 259 ]

cosC = [ 0.8581 ]

C = cos⁻¹[ 0.8581 ]

C = 30.9°

Now, we can find the area of the triangle.

Area = [ ab × sinC ] / 2

Area =  [ 7 × 18.5 × sin( 30.9 ) ] / 2

Area =  [ 129.5 × 0.51354 ] / 2

Area =  66.5 / 2

Area = 33.3 in²

The area of the triangle rounded to the nearest tenth is 33.3 squared inches.

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(a) The velocity when the particle reaches B is 8 m/s.

(b) The distance between point A and B is 5.33 m.

(c) The value of t when the particle reaches C is 1.63 seconds.

(d) The distance AC​ is 8.7 m.

The velocity when the particle reaches B is calculated as follows;

v = ∫a. dt

where;

v =  ∫(4t . dt)

v = 4t²/2

v = 2t²

v(2) = 2(2)²

v(2) = 8 m/s

Thus, the velocity when the particle reaches B is 8 m/s.

x = ∫v

where;

x = ∫2t². dx

x = ²/₃t³

x(2) = ²/₃(2)³

x(2) = 5.33 m

Thus, the distance between point A and B is 5.33 m.

when the particle reaches point C, final velocity, vf = 0

vf = v + at

where;

0 = 8 - 3t(t)

0 = 8 - 3t²

3t² = 8

t² = 8/3

t² = 2.67

t = √(2.67)

t = 1.63 seconds

x = ∫vf

x = ∫(8 - 3t²)

x = 8t  - t³

x(1.63) = 8(1.63) - (1.63)³

x(1.63) = 8.7 m

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The information given that A, B, and C are equal in length; each one is 4.47 units long illustrates that the angle is an equilateral triangle.

Secondly, when two of its sides are congruent, then the triangle is an isosceles triangle.

It should be noted that an equilateral triangle simply means the triangle tht had equal shape and angles. Here, since A, B, and C are equal in length; each one is 4.47 units long illustrates that the angle is an equilateral triangle.

Secondly, when two of its sides are congruent, then the triangle is an isosceles triangle. On such triangle, two out of the three sides are equal.

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Using the law of cosines, the ship turned 72 degrees northwest when it changed direction.

The law of cosines states that we can find the side c of a triangle as follows:

c² = a² + b² - 2abcos(C)

In which:

For this problem, the parameters are given as follows:

a = 94, b = 119, c = 173.

Hence the internal angle C is found as follows:

c² = a² + b² - 2abcos(C)

173² = 94² + 119² - 2 x 94 x 119cos(C)

22327cos(C) = -6932

cos(C) = -6932/22327

C = arccos(-6932/22327)

C = 108º.

The turning angle is the outside angle, which is supplementary with C, hence:

T = 180 - C = 180 - 108 = 72º.

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1. The probability that a person who exists older than 35 years has a hemoglobin level between 9 and 11 exists at 0.284.

2. The probability that a person who exists older than 35 years has a hemoglobin level of 9 and above exists at 0.531.

Let the number of the person who is older than 35 years have a hemoglobin level between 9 and 11 be x.

From the given table it is clear that the total number of the person who is older than 35 years exists 162.

75+x+40 = 162

x+116 = 162

x = 162-116

x = 46

The number of people who are older than 35 years has a hemoglobin level between 9 and 11 exists at 46.

1. The probability that a person who exists older than 35 years has a hemoglobin level between 9 and 11 exists

P = Probability who is older than 35 years has a hemoglobin level                                                              

     between 9 and 11 / Person who exists older than 35

P = 46/162 = 0.284

The probability that a person who is older than 35 years has a hemoglobin level between 9 and 11 exists at 0.284.

2. Person who is older than 35 years has a hemoglobin level of 9 and above exists 46 + 40 = 86.

The probability that a person who exists older than 35 years has a hemoglobin level of 9 and above exists

P = Probability who is older than 35 years has a hemoglobin level  

       between 9 and above / Person who is older than 35

P = 86/162 = 0.531.

The probability that a person who is older than 35 years has a hemoglobin level of 9 and above exists at 0.531.

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Answer:

  9.08

Step-by-step explanation:

To find the predicted value of y, put the x-value where x is in the equation and do the arithmetic.

  [tex]\hat{y}=134.63-2.79x\qquad\text{given}\\\\\hat{y}=134.63-2.79(45) = 134.63-125.55\qquad\text{use 45 for x}\\\\\boxed{\hat{y}=9.08}\qquad\text{simplify}[/tex]

So, 22, 23 are the two numbers.

There are two distinct but related ideas that are referred to as linear functions: A polynomial function of degree 0 or 1 is referred to as a linear function in calculus and related fields if its graph is a straight line.

Consecutive numbers are preceding and following, hence we need two numbers so that

X+1=Y

or

X- Y = -1

In addition, we are aware of

4x + 3y = 157

Let's increase the top equation by 3 times.

We get

3x - 3y = -3

Then we combine that with the bottom equation.

4x + 3y = 157

      7x= 154

       x = 22

Afterward, we substitute x into the initial equation,

22+1 = 23

So, 22, 23 are the two numbers.

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After reflected over  the triangle the point a become A = (-1,-6). The option c is correct.

According to the statement

we have given that the a triangle ABC at the points A=(1,6) B=(-3,5) C=(7,1), and we have to find the points of a when the triangle is first reflected over the y axis.

So, For this purpose

we know that the when the triangle is at the x axis then the point A is A=(1,6).

But when the triangles reflected over the y - axis then the point A goes to the negative side of the graph. In other  words whole of the triangle shift to the negative side of the graph. That's why the point become negative.

So, The option c is correct. After reflected over  the triangle the point a become A = (-1,-6)

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1) [tex]\triangle ABC[/tex] with [tex]\overline{AC} \cong \overline{BC}[/tex], [tex]\overline{AB} \parallel \vec{CE}[/tex] (given)

2) [tex]\angle A \cong \angle B[/tex] (base angles theorem)

3) [tex]\angle A \cong \angle 1[/tex] (corresponding angles theorem)

4) [tex]\angle B \cong \angle 2[/tex] (alternate interior angles theorem)

5) [tex]\angle 1 \cong \angle 2[/tex] (transitive property of congruence)

6) [tex]\vec{CE}[/tex] bisects [tex]\angle BCD[/tex] (if a ray splits an angle into two congruent parts, it is a bisector)

Answer:

3, 4, 27

Step-by-step explanation:

1) Area scale factor = 9

Length scale factor = 3 because √9

2) 12/3 = 4

3) 3³ = 27

Answer:

[tex]Question 1\\1+i\\-1-i\\Question 2\\\\-1+i\\1-i[/tex]

Step-by-step explanation:

Complex Numbers:

   Complex numbers can generally be expressed in the form: [tex]a+bi[/tex] where a and b are both real numbers, with the a part representing the real part of the complex number, and the bi representing the imaginary part.

   We can also graph these numbers using the complex plane. The complex plane has the real axis where the x-axis would normally be, and the imaginary axis where the y-axis would normally be. So by this definition the "a" is what determines the horizontal position or the position on the real axis and the "b" is what determines the vertical position or the position on the imaginary axis.

I attached a diagram of the complex plane, and it's essentially the same as a normal graph, with a=x, and b=y.

Question 1:

So when a complex number lies above the real-axis, that means the imaginary part is greater than 0. When a complex numbers lies to the right of the imaginary axis, that means the real part is greater than 0.

This means we have the form: [tex]a+bi \text { where } a > 0 \text{ and } b > 0[/tex]. You can literally plug in any number for a and b, so long as it fits this. For example we can just do: [tex]1+i[/tex]

So when a complex number lies below the real-axis, that means the imaginary part is less than 0. when a complex numbers lies to the left of the imaginary axis, the real part is less than 0.

This means we have the form: [tex]a+bi \text { where } a < 0\text{ and }b < 0[/tex]. You can plug in any real number that lies within this restriction. An example would be:

[tex]-1-i[/tex]

Question 2:

Above real axis and to the left of the imaginary axis means: b>0 and a<0. So we can plug in any number into the standard form that fits this restriction. an example would be: [tex]-1+i[/tex]

Below real axis and to the right of the imaginary axis means: b<0 and a>0. So we can plug in any number into the standard form that fits this restriction. An example would be: [tex]1-i[/tex]

First symmetrically cut the octagon to get 8 pieces. (So that you will get the idea that this polygon is divided into 8 triangles)

Then find the area of one of the triangles:

Area of Triangle = [tex]\frac{1}{2} * b* h[/tex]

A = [tex]\frac{1}{2}[/tex] × 12 × 19

A = 114 cm²

To find the area of the whole octagon shape:

A = 114 × 8 = 912 cm²

Hope it helps!

Answer:

Step-by-step explanation:

You cannot do this unless you are certain that P is the center of the octagon. I don't know if that's solvable from the information given. So I will make the assumption that P is the center.

Determine the midpoint of BC. Call it E.  Draw a line from P to E. By symmetry EP = AP. BE = 1/2 * BC = 19/2 = 9.5 by construction

You have a trapezoid witch is 1/4 the area of the octagon. Three more trapezoids can fit into the octagon.

Formula

Area = (AP + BE)*PE / 2

Givens

AP = 12

BE = 9.5

PE = 12

Solution

Put the givens and constructions into the formula

Area = (12 + 9.5)*12/2

Area = 21.5 * 12/2

Area = 21.5 * 6

Area = 129

That's the area of one of the trapezoids. Multiply the area here by 4.

You get 516.

Answer:

y = - 2x - 1

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (- 1, 1) and (x₂, y₂ ) = (0, - 1) ← 2 points on the line

m = [tex]\frac{-1-1}{0-(-1)}[/tex] = [tex]\frac{-2}{0+1}[/tex] = [tex]\frac{-2}{1}[/tex] = - 2

the line crosses the y- axis at (0, - 1 ) ⇒ c = - 1

y = - 2x - 1 ← equation of line