The sum of the lengths of two opposite sides of the circumscribed quadrilateral is 12 cm, the length of a radius of the circle is 2 cm. Find the area of the quadrilateral.

Answer:

  C. Yes; Point B prevents this function from being linear because (12, -16) would only satisfy a function with a variable rate of change.

Step-by-step explanation:

Point B is not on the line shown in the graph, so the function would have to be non-linear to include point B.

Answer:

[tex]T.S.A=40\pi in^2[/tex]

Step-by-step explanation:

The total surface area of the cone is

[tex]\pi r^2+\pi rl[/tex]

The radius of the cone is 4 inches.

The slant height of the cone is l=6 inches.

We substitute these values into the formula to obtain;

[tex]T.S.A=\pi \times (4)^2+\pi \times 4\times6[/tex]

[tex]T.S.A=16\pi+24\pi[/tex]

[tex]T.S.A=40\pi in^2[/tex]

Or

[tex]T.S.A=125.7in^2[/tex]  to the nearest tenth.

Answer:

a. 215.6 in^3

b. 1.51 lb

Step-by-step explanation:

The area of each hand grip hole is that of a circle of radius 0.6 in together with a rectangle 2 in long and 1.2 in wide. So, that area is ...

π·(0.6 in)^2 + (2 in)(1.2 in) = (0.36π +2.4) in^2

The area of the kickboard before the hand grip holes are put in is that of a semicircle of radius 5.5 in together with a rectangle 12 in long and 11 in wide. So, that area is ...

(1/2)·π·(5.5 in)^2 + (12 in)(11 in) = (15.125π +132) in^2

Taking the hand grip holes out, the top area of the board is ...

((15.125π +132) -2(0.36π +2.4)) in^2

= (14.405π + 127.2) in^2

___

a. The volume is the product of the area and the thickness, so is ...

((14.405π +127.2) in^2)·(1.25 in) ≈ 215.568 in^3

__

b. The weight of the kickboard is the product of its volume and its density:

(215.568 in^3)(0.007 lb/in^3) ≈ 1.509 lb

To find the volume of the foam kickboard, its length, width, and height are needed. The weight is then calculated by multiplying the volume by the weight per cubic inch. Without specific dimensions, we cannot provide exact answers for the volume and weight.

To find the volume of the foam kickboard, we would need its dimensions such as length, width, and height. If we had these measurements, the volume (V) can be calculated using the formula V = length × width × height. Unfortunately, the question does not provide specific dimensions, so we cannot calculate an exact volume without this information.

To calculate the weight of the kickboard, once the volume is determined, you would multiply the volume by the weight per cubic inch of the foam. Assuming we had a volume of V cubic inches, the weight (W) of the kickboard can be found with W = V × 0.007 lb/in3.

For example, if the kickboard's volume was 100 cubic inches, then the weight would be 100 × 0.007 lb/in3 = 0.7 lb.

Answer:

Step-by-step explanation:

So we're looking at a rectangle split into 5 smaller rectangles.  If the height of each rectangle is y and the width of each rectangle is x, then the amount of fencing is:

P = 6y + 10x

And the area of the large rectangle is:

A = 5xy

We know that P = 780:

780 = 6y + 10x

10x = 780 - 6y

5x = 390 - 3y

If we substitute this into our area equation:

A = (390 - 3y) y

A = -3y² + 390y

This is a vertical parabola pointing down, so we know the maximum is at the vertex, which is at -b/(2a).  Or, we can use calculus to take the derivative and set to 0.

dA/dy = -6y + 390

0 = -6y + 390

y = 65

Solving for x:

5x = 390 - 3y

5x = 390 - 3(65)

5x = 195

x = 39

So each part will have a width of 39 feet and a height of 65 feet.

I guess the function is [tex]f(x)=9-x^{2/3}[/tex]. Then [tex]f(-27)=0[/tex] and [tex]f(27)=0[/tex].

The derivative is [tex]f'(x)=-\dfrac23 x^{-1/3}[/tex], but there is no [tex]c[/tex] such that

[tex]-\dfrac23c^{-1/3}=0[/tex]

This doesn't contradict Rolle's theorem because [tex]f'(0)[/tex] does not exists. In other words, [tex]f[/tex] is not differentiable on (-27, 27), so the conditions of Rolle's theorem are not met. (Looks like that would be the last option, or the second to last option if the last one is "Nothing can be concluded")

The function f(x) = 9 - x²/3 is even, and its derivative f'(c) equals to zero at c=0, which lies within the interval (-27, 27). Therefore, it does not contradict Rolle's theorem.

The function in question is f(x) = 9 - x²/3. When we substitute x with -27 and 27, we get f(-27) = 9 - ((-27)²/3) = -243 and f(27) = 9 - (27²/3) = -243. This confirms that the function is even as f(-27) = f(27).

To find the critical values, we'll take the derivative of the function, which gives us f'(x) = -2x/3. We set f '(c) = 0, solving for c, and determine c = 0. Rolle's Theorem states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the interval (a, b) such that f '(c) = 0. With f(-27) = f(27) and the derivative proving to be zero at c=0 (which is inside the interval (-27,27)) this does not contradict Rolle's theorem.

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Answer with explanation:

Given:

In Δ DEF, ∠3=∠4.

To prove:→ DE=E F

Proof:

1.  ∠3=∠4------[Given]

2. →∠1 and ∠ 3 are Supplementary to each other.

(a)⇒∠1 + ∠ 3=180°

 →∠2 and ∠ 4 are also Supplementary to each other.

(b)⇒∠2 + ∠ 4=180°

                                     --------------------[Exterior sides in opposite rays]

3. From 1 , a and b

 ⇒∠ 1 = ∠ 2-------[Two Angles Supplementary to equal Angles are equal to each other]

4. [tex]\Bar{DE}=\Bar{E F}[/tex]

 If two angles of a Triangle are equal , then side opposite to these angles are equal.

Answer:

Step-by-step explanation:

Answer:

y = -2(x -1)^2 -2

Step-by-step explanation:

My "work" consists of providing a table of values to a calculator and asking it for a quadratic model. The result is ...

y = -2(x -1)^2 -2

__

If you like to do these "by hand", you can write the model, then solve for its parameters using the given points.

We observe that the first and third points have the same y-coordinate. Then the vertex of the quadratic will be halfway between the corresponding x-values, at ...

h = (-2 +4)/2 = 1

So, one of the parameters of the model is found already. Using the second point and one other, we can find the remaining parameters for our model:

y = a(x -1)^2 +k

for (4, -20) ...

-20 = a(4 -1)^2 +k = 9a +k

for (0, -4) ...

-4 = a(0 -1)^2 +k = a + k

Subtracting the second equation from the first, we get

-16 = 8a

-2 = a . . . . . divide by 8

Substituting this value of a into the second equation, we have ...

-4 = -2 +k

-2 = k . . . . . . add 2

So, our model is ...

y = -2(x -1)^2 -2

Answer:

Exponential

Step-by-step explanation:

i took that

Answer:

2

Step-by-step explanation:

count how many more

Answer is letter A

it is the answer

For this case we must find the difference of the following expressions:

[tex]0.00067\\2.3 * 10 ^ {- 5}[/tex]

For the second expression we must run the decimal 5 times to the left, because the exponent is negative, that is:

[tex]0.000023[/tex]

We subtract:

[tex]0.00067-0.000023 = 0.000647[/tex]

Represented in scientific notation we have:

[tex]6.47 * 10 ^ {- 4}[/tex]

Answer:

Option A

Answer:

a=4/5y

Step-by-step explanation:

Answer:

it is $30.00

Step-by-step explanation:

C. 4

First, rearrange the equation into slope intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. You get y = -5x + 4. This means the y-intercept is 4.

For this case we have a function of the form [tex]y = f (x)[/tex]

Where:

[tex]f (x) = 4-5x[/tex]

To find the y-intercept of the function we must do x = 0.

Then, replacing:

[tex]f (0) = 4-5 (0)\\f (0) = 4-0\\f (0) = 4[/tex]

So, the y-intercep of the function is 4

ANswer:

4

Option C

Answer:

  c)  110°

Step-by-step explanation:

Arc FP is twice the measure of the marked angle, so is 70°. If FD is supposed to be a diameter, then arc FPD is a semicircle (180°) and arc PD is 180° -70° = 110°.

In technical translation, 4 x 3 - 1 is less than or equal to 11 (it's equal). 4 x 3 - 1 < 11 is not true because 11 is not less than 11.

Hope this helps!

Explaining why x = 3 satisfies 4x − 1 ≤ 11 but not 4x − 1 < 11:

When x = 3, we can evaluate the inequalities:

For 4x − 1 ≤ 11: 4(3) - 1 ≤ 11, which simplifies to 12 ≤ 11, making it true.

For 4x − 1 < 11: 4(3) - 1 < 11, which simplifies to 12 < 11, making it false.

Therefore, when x = 3, the first inequality is true while the second one is false.

Answer:

m=13 b=17

Step-by-step explanation:

Answer:

m=13 b=17

hope this helps

Number 2,3,5 are true

5 x tan 66 = 7.140

your welcome :)

Answer:

Step-by-step explanation:

Answer:

B. (1,7)

Step-by-step explanation:

Answer is B. (1,7)

If x = 1 then

y > 1 + 5

7 > 6

Answer:

(1 , 7) is a solution of y > IxI + 5 ⇒ answer B

Step-by-step explanation:

* Lets revise the absolute value

- IxI = positive value

- IxI can not give negative value

- The value of x could be positive or negative

* Lets solve the problem

∵ y > IxI + 5

∴ y > x + 5 OR y > -x + 5

- Lets check the answers

∵ y > 0 + 5 ⇒ y > 5

- But y = 5, and 5 it is not greater than 5 and there is no difference

 between the two cases because zero has no sign

∴ (0 , 5) not a solution

∵ y > 1 + 5 ⇒ y > 6

- Its true y = 7 and 7 is greater than 6

∵ y > -1 + 5 ⇒ y > 4

- Its true y = 7 and 7 is greater than 4

∴ (1 , 7) is a solution

∵ y > 7 + 5 ⇒ y > 12

- But y = 1 and 1 is not greater than 12

∵ y > -7 + 5 ⇒ y > -2

- Its true y = 1 and 1 is greater than -2

* we can not take this point as a solution because it is wrong

 with one of the two cases

∴ (7 , 1) is not a solution

Answer:

  8 square units

Step-by-step explanation:

The figure is a trapezoid. The area of it is given by the formula ...

  A = (1/2)(b1 +b2)h

where b1 and b2 are the lengths of the parallel bases and h is the distance between them.

Your figure shows the base lengths to be 5 and 3, and their separation to be 2. Filling the numbers in the formula, we have ...

  A = (1/2)(5 +3)(2) = (1/2)(8)(2) = 4·2 = 8

The area of the figure is 8 square units.

_____

The right-pointing arrows on the horizontal lines identify those lines as being parallel. The right-angle indicator and the 2 next to the dotted line indicate the perpendicular distance between the parallel lines is 2 units.

Answer:

The point (2 - d, y) is on the graph of f(x)

The line x = 2 is the axis of symmetry of the graph of f(x)

Step-by-step explanation:

* Lets explain how to prove that a point lies on a graph Algebraically

- Substitute the value of the x-coordinate of the point in the equation

 of the graph the answer must be equal the y-coordinate of the point

- The function is a quadratic because the greatest power of x is 2,

  then it represented by parabola

- The parabola has a vertex point (h , k), where h is the x-coordinate

  and k is the y-coordinate

- This vertex divides the parabola into two equal parts, then the axis

 of symmetry of the parabola is a vertical line passing through it

∴ The equation of the axis of symmetry is x = h

- The vertex of the parabola could be minimum point if the parabola

  opened upward or maximum if it opened downward

- The minimum value and the maximum value are the value of k

# Look to the attached figures for more understand

* Now lets solve the problem

∵ f(x) = x(x - 4)

∵ Point (2 + d , y) is on the graph of f(x)

- Replace each x in f(x) by 2 + d

∴ f(2 + d) = (2 + d)(2 + d - 4) ⇒ add 2 and -4

∴ f(2 + d) = (2 + d)(-2 + d)

∵ f(2 + d) = y

∴ y = (2 + d)( -2 + d)

* Multiply them to simplify

∴ y = 2(-2) + 2(d) + d(-2) + d(d) = -4 + 2d - 2d + d²

∴ y = -4 + d²

* Lets do these steps again with point (2 - d , y)

- Replace each x in f(x) by 2 - d

∴ f(2 - d) = (2 - d)(2 - d - 4) ⇒ add 2 and -4

∴ f(2 - d) = (2 - d)(-2 - d)

∵ f(2 - d) = y

∴ y = (2 - d)( -2 - d)

* Multiply them to simplify

∴ y = 2(-2) + 2(-d) - d(-2) - d(-d) = -4 - 2d + 2d + d²

∴ y = -4 + d²

- The value of y of the point (2 - d , y) = the value of y of the point on

  the graph

∵ f(2 + d) = f(2 - d)

∵ The point (2 + d , y) is on the graph of f(x)

∴ The point (2 - d , y) is on the graph of f(x)

* It is true the point (2 - d, y) is also on the graph.

* To find the relation between the line x = 2 and the graph of f(x)

  lets find the vertex of the parabola

- If f(x) = ax² + bx + c in the general form, where a, b , c are constant

 then h = -b/2a, where h is the x-coordinate of the vertex point, a is

 the coefficient of x² and b is the coefficient of x

∵ f(x) = x(x - 4) ⇒ multiply the bracket by x to put it in the general form

∴ f(x) = x² - 4x

- Find the value of a and b to find h

∵ a = 1 and b = -4

∵ h = -b/2a

∴ h = -(-4)/2(1) = 4/2 = 2

∴ The x-coordinate of the vertex point = 2

∵ The axis of symmetry of the parabola passing through the

   vertex point

∴ The equation of the axis of symmetry of the parabola is x = 2

* The line x = 2 is the axis of symmetry of the graph of f(x)

Answer:

Step-by-step explanation:

AYOOO

Step-by-step explanation:

it is solved in the diagram

Answer:

  A  404 mi

Step-by-step explanation:

If we designate the points of the triangle A, B, and C for the locations of the western station, eastern station, and balloon, respectively, we have the following:

  ∠CAB = 90° - 41° = 49°

  ∠CBA = 90° + 21° = 111°

  ∠ACB = 41° -21° = 20°

side "c" (opposite ∠ACB) is 148 miles

The distance we're asked to find is AC = b, the longest side of the triangle. The law of sines tells us ...

  b/sin(B) = c/sin(C)

  b = c·sin(B)/sin(C) = (148 mi)·sin(111°)/sin(20°) ≈ 403.98 mi ≈ 404 mi

Using trigonometry and the law of sines, the distance from the balloon to the western station is approximately 373 miles.

This is a problem involving trigonometry, particularly the use of the law of sines. The two tracking stations and the balloon form a triangle. The angle at the western station is 41°, the angle at the eastern station is (180 - 21 - 41) = 118°, and the distance between the two stations (the side opposite to the 41° angle) is 148 miles. According to the Law of Sines, the ratio of each side of the triangle to the sine of its opposite angle is constant. Thus, we can set up the equation sin(41°) / x = sin(118°) / 148 miles, where x represents the distance from the balloon to the western station. Solving for x gives us approximately 373 miles.

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Answer:

[tex]t=2.47\ s[/tex]  

Step-by-step explanation:

The equation that models the height of the ball in feet as a function of time is

[tex]h(t) = h_0 + s_0t -16t ^ 2[/tex]

Where [tex]h_0[/tex] is the initial height, [tex]s_0[/tex] is the initial velocity and t is the time in seconds.

We know that the initial height is:

[tex]h_0 = 4.5\ ft[/tex]

The initial speed is:

[tex]s_0 = 110sin(20\°)\\\\s_0 = 37.62\ ft/s[/tex]

So the equation is:

[tex]h (t) = 4.5 + 37.62t -16t ^ 2[/tex]

The ball hits the ground when when [tex]h(t) = 0[/tex]

So

[tex]4.5 + 37.62t -16t ^ 2 = 0[/tex]

We use the quadratic formula to solve the equation for t

For a quadratic equation of the form

[tex]at^2 +bt + c[/tex]

The quadratic formula is:

[tex]t=\frac{-b\±\sqrt{b^2 -4ac}}{2a}[/tex]

In this case

[tex]a= -16\\\\b=37.62\\\\c=4.5[/tex]

Therefore

[tex]t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}[/tex]

[tex]t_1=-0.114\ s\\\\t_2=2.47\ s[/tex]  

We take the positive solution.

Finally the ball takes 2.47 seconds to touch the ground

Answer:

  (-2, 7) and (0, 3)

Step-by-step explanation:

A graph of the two equations clearly shows the points of intersection.

The equations are conveniently graphed by a graphing calculator (as here) or by a spreadsheet program, on-line graphing tool, or graphing app.

___

Alternate solution

You can set the two values of y equal to each other, then solve for x.

  3x^2 +4x +3 = -2x +3

  3x^2 +6x = 0 . . . . . subtract the right side expression

  3(x)(x +2) = 0 . . . . . factor the equation

  x = 0, x = -2 . . . . . . solutions that make the factors zero

  y = -2{0, -2} +3 . . . . substitute the values of x into the expression for y

  y = {0, 4} +3

  y = {3, 7} . . . . . . . . . the values of y corresponding to x = {0, -2}

Then the points of intersection are (0, 3) and (-2, 7).

Answer:

rate of the plane in still air is 33 miles per hour and the rate of the wind is 11 miles per hour

Step-by-step explanation:

We will make a table of the trip there and back using the formula distance = rate x time

                 d              =            r       x       t

there

back

The distance there and back is 264 miles, so we can split that in half and put each half under d:

               d            =         r        x        t

there    132

back     132

It tells us that the trip there is with the wind and the trip back is against the wind:

              d           =        r         x        t

there     132        =    (r + w)

back      132       =     (r - w)

Finally, the trip there took 3 hours and the trip back took 6:

             d         =        r        *        t

there    132     =    (r + w)     *       3

back     132     =    (r - w)      *       6

There's the table.  Using the distance formula we have 2 equations that result from that info:

132 = 3(r + w)        and        132 = 6(r - w)

We are looking to solve for both r and w.  We have 2 equations with 2 unknowns, so we will solve the first equation for r, sub that value for r into the second equation to solve for w:

132 = 3r + 3w and

132 - 3w = 3r so

44 - w = r.  Subbing that into the second equation:

132 = 6(44 - w) - 6w and

132 = 264 - 6w - 6w and

-132 = -12w so

w = 11

Subbing w in to solve for r:

132 = 3r + 3(11) and

132 = 3r + 33 so

99 = 3r and

r = 33

Answer: Second Option

[tex]f (x) = 2 | x-2 | +6[/tex]

Step-by-step explanation:

For a function of the form:

[tex]f (x) = a | x-h | + k[/tex]

The vertex is always at the point (h, k)

In this case we know that the vertex is in the point (2, 6) and [tex]a = 2[/tex]

This means that

[tex]h = 2\\\\k = 6[/tex]

Therefore the function that has its vertivce in the point (2, 6) is:

[tex]f (x) = 2 | x-2 | +6[/tex]

The correct answer is the second

Answer:

Part 1) The measure of angle EYL is [tex]96\°[/tex]

Part 2) The measure of arc EHL is [tex]108\°[/tex] and the measure of arc LVE is [tex]252\°[/tex]

Step-by-step explanation:

we know that

The measurement of the outer angle is the semi-difference of the arcs which comprises

Part 1)

Let

x------> the measure of arc LHE

y----> the measure of arc LVE

we know that

[tex]x+y=360\°[/tex]

[tex]x=84\°[/tex]

Find the value of y

[tex]y=360\°-84\°=276\°[/tex]

Find the measure of angle EYL

[tex]m<EYL=\frac{1}{2} (y-x)[/tex]

substitute the values

[tex]m<EYL=\frac{1}{2}(276\°-84\°)=96\°[/tex]

Part 2)

Let

x------> the measure of arc EHL

y----> the measure of arc LVE

we know that

[tex]x+y=360\°[/tex]

[tex]x=360\°-y[/tex] -----> equation A

[tex]m<EYL=72\°[/tex]

[tex]m<EYL=\frac{1}{2} (y-x)[/tex]

substitute

[tex]72\°=\frac{1}{2} (y-x)[/tex]

[tex]144\°=(y-x)[/tex]

[tex]x=y-144\°[/tex] --------> equation B

equate equation A and equation B and solve for y

[tex]360\°-y=y-144\°[/tex]

[tex]2y=360\°+144\°[/tex]

[tex]2y=504\°[/tex]

[tex]y=252\°[/tex]

Find the value of x

[tex]x=252\°-144\°=108\°[/tex]

therefore

The measure of arc EHL is [tex]108\°[/tex]

The measure of arc LVE is [tex]252\°[/tex]

A circle is a curve sketched out by a point moving in a plane. The measure of the arcEHL and arcLVE are 108° and 252° respectively.

A circle is a curve sketched out by a point moving in a plane so that its distance from a given point is constant; alternatively, it is the shape formed by all points in a plane that are at a set distance from a given point, the centre.

As we can see in the image attached below, the radius of the circle is OE and OL while there is two tangent to the circle Ey and LY. Therefore, the measure of the ∠OEY and ∠OLY is 90°.

A.)

The sum of the angles of a quadrilateral is 360°.

As it is mentioned in the problem the measure of the angle made by arc LHE is 84° which means the measure of ∠EOL is 84°. Therefore, the sum of all the angles can be written as,

[tex]\text{Sum of angles} = 360^o\\\\\angle OEY + \angle OLY + \angle EOL + \angle EYL = 360^o\\\\90^o+90^o+84^o +\angle EYL = 360^o\\\\\angle EYL = 96^o[/tex]

B.)

The sum of the angles of a quadrilateral is 360°.

As it is mentioned in the problem the measure of ∠EYL is 72°. Therefore, the sum of all the angles can be written as,

[tex]\text{Sum of angles} = 360^o\\\\\angle OEY + \angle OLY + \angle EOL + \angle EYL = 360^o\\\\90^o+90^o+ \angle EOL +72^o = 360^o\\\\ \angle EOL= 108^o\\\\ \rm arc EHL=108^o[/tex]

Since a complete circle measures 360°, therefore, the sum of the angles made by arc EHL and arc LVE can be written as,

arcEHL + arcLVE = 360°

108° + arcLVE = 360°

arcLVE = 252°

Thus, the measure of the arcEHL and arcLVE are 108° and 252° respectively.

Learn more about Circle:

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X1= -3/2

X2= 0

F(x)=4x^2+6x

To find X-intercept/zero, substitute f(x)=0

0=4x^2+6x

Move the constant to the right

4x^2+6x=0

Factor out 2x from the expression

2x(2x+3)=0

Divide both sides of the equation

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

X(2x+3)=0

When the product of factors equals 0, at least one factor is 0.

X=0

2x+3=0

Next you solve for X by moving the constant to the right

2x=-3

Then divide both sides by 2

X=-3/2

Hope this answers your question.

The result is (f/g)(x) = (4x² + 6x) / (2x² + 13x + 15).

To find (f/g)(x) for the given functions f(x) = 4x² + 6x and g(x) = 2x² + 13x + 15, you need to divide the function f(x) by g(x).

Step-by-Step Solution:

Therefore, (f/g)(x) stands as (4x² + 6x) / (2x² + 13x + 15).

Answer:D pie

Step-by-step explanation:

You get 2pie/2 cancel out you get pie

The period of y=cos(2x) is π, so the frequency is 1/π.

The key features of the graph are:

Amplitude: The amplitude of the graph is 1, which means the function oscillates between −1and 1.

Period: The period of the graph is π, which means the function completes one cycle every πunits on the x -axis. This is because the frequency of the function is 2, which means it completes two cycles for every 2π units on the x -axis.

Midline: The midline of the graph is y=0. This is because the function is neither shifted up nor down.

Extrema: The graph has maxima at x=2kπ​ for any integer k, and minima at x=2(2k+1)π​for any integer k.

The graph of the function y=cos(2x). The frequency of a trigonometric function is the reciprocal of its period. The period of y=cos(2x) is π, so the frequency is 1/π.