a practice field is 250 ft long the game is 40% longer than the practice field of how long is the game

n order to bring back customers to the AMC Movie Theaters, they are offering two movie plans; a Family Plan of 4 used to cost $2

7nadin3 [17]

Is gonna be the family plan please give brainlist and help me report the comment on top is a page virus

4 0

3 years ago

Read 2 more answers

Write an equation for the perpendicular bisector of the line segment joining the two points.

Leona [35]

Points (1, 7) and (-3, 2)

Slope for a line between (x₁, y₁) and (x₂, y₂) , m = (y₂ -y₁) / (x₂- x₁)

The slope for the line joining the two points = (2 - 7) / (-3 - 1) = -5/-4

Slope = 5/4

Hence the perpendicular bisector would have a slope of -1/(5/4) = -4/5

By condition of perpendicularity

For points (1, 7) and (-3, 2),

Formula for midpoints for (x₁, y₁) and (x₂, y₂) is ((x₁ +x₂)/2 , (y₁+ y₂)/2)

Midpoint for (1, 7) and (-3, 2) = ((1+ -3)/2 , (7+2)/2) = (-2/2, 9/2)

= (-1, 9/2)

Since the slope of perpendicular bisector is -4/5 and passes through the midpoint (-1, 9/2)

Equation y - y₁ = m (x - x₁)

y - 9/2 = (-4/5) (x - -1)

y - 9/2 = (-4/5)(x + 1)

   5(y - 9/2) = -4(x + 1)

   5y - 45/2 = -4x - 4

   5y = -4x - 4 + 45/2

5y + 4x = 45/2 - 4

5y + 4x = 22 1/2 - 4 = 18 1/2

5y + 4x = 37/2

10y + 8x = 37

The equation of the line to perpendicular bisector is 10y + 8x = 37

7 0

4 years ago

A group of friends were working on a student film that had a budget of $1300. They used $546 of their budget on props. What perc

Katarina [22]

To find percentage: divide the given value by the total and multiply by 100

546/1300 x 100 = 42%

5 0

2 years ago

In Exercises 11-18, use analytic methods to find the extreme values of the function on the interval and where they occur. Identi

Colt1911 [192]

Answer:

Absolute maximum of 1 at x = pi/4 ; $(\frac{\pi}{4}, \ 1)$

Absolute minimum of -1 at x = 5pi/4 ; $(\frac{5\pi}{4} , \ -1)$

Local maximum of √2/2 at x = 0 ; $(0, \ \frac{\sqrt{2} }{2} )$

Local minimum of 0 at x = 7pi/4 ; $(\frac{7\pi}{4}, \ 0)$

No critical points that are not stationary points.

Step-by-step explanation:

$f(x)=sin(x+\frac{\pi}{4} ), \ 0 \leq x\leq \frac{7 \pi}{4}$

<h2>Take Derivative of f(x):</h2>

Let's start by taking the derivative of the function.

Use the power rule and the chain rule to take the derivative of f(x).

$f'(x)=\frac{d}{dx} [sin(x+\frac{\pi}{4})] \times \frac{d}{dx} (x+\frac{\pi}{4})$

The derivative of sin(x) is cos(x), so we can write this as:

$f'(x)=cos(x+\frac{\pi}{4})\times \frac{d}{dx} (x+\frac{\pi}{4})$

Now, we can apply the power rule to x + pi/4.

$f'(x)=cos(x+\frac{\pi}{4} ) \times 1$

$f'(x)=cos(x+\frac{\pi}{4} )$

<h2>Critical Points: Set f'(x) = 0</h2>

Now that we have the first derivative of $f(x)=sin(x+\frac{\pi}{4})$ , let's set the first derivative to 0 to find the critical points of this function.

$0=cos(x+\frac{\pi}{4})$

Take the inverse cosine of both sides of the equation.

$cos^-^1(0) = cos^-^1[cos(x+\frac{\pi}{4})]$

Inverse cosine and cosine cancel out, leaving us with x + pi/4. The inverse cosine of 0 is equal to 90 degrees, which is the same as pi/2.

$\frac{\pi}{2} = x +\frac{\pi}{4}$

Solve for x to find the critical points of f(x). Subtract pi/4 from both sides of the equation, and move x to the left using the symmetric property of equality.

$x=\frac{\pi}{2}- \frac{\pi}{4}$

$x=\frac{2 \pi}{4}-\frac{\pi}{4}$
$x=\frac{\pi}{4}$

Since we are given the domain of the function, let's use the period of sin to find our other critical point: 5pi/4. This is equivalent to pi/4. Therefore, our critical points are:

$\frac{\pi}{4}, \frac{5 \pi}{4}$

<h2>Sign Chart(?):</h2>

Since this is a sine graph, we don't need to create a sign chart to check if the critical values are, in fact, extreme values since there are many absolute maximums and absolute minimums on the sine graph.

There will always be either an absolute maximum or an absolute minimum at the critical values where the first derivative is equal to 0, because this is where the sine graph curves and forms these.

Therefore, we can plug the critical values into the original function f(x) in order to find the value at which these extreme values occur. We also need to plug in the endpoints of the function, which are the domain restrictions.

Let's plug in the critical point values and endpoint values into the function f(x) to find where the extreme values occur on the graph of this function.

<h2>Critical Point Values:</h2>

$f(\frac{\pi}{4} )=sin(\frac{\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{\pi}{4} )=sin(\frac{2\pi}{4}) \\ f(\frac{\pi}{4} )=sin(\frac{\pi}{2}) \\ f(\frac{\pi}{4} )=1$

There is a maximum value of 1 at x = pi/4.

$f(\frac{5\pi}{4} )=sin(\frac{5\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{5\pi}{4} )=sin(\frac{6\pi}{4}) \\ f(\frac{5\pi}{4}) = sin(\frac{3\pi}{2}) \\ f(\frac{5\pi}{4} )=-1$

There is a minimum value of -1 at x = 5pi/4.

<h2>Endpoint Values:</h2>

$f(0) = sin((0) + \frac{\pi}{4}) \\ f(0) = sin(\frac{\pi}{4}) \\ f(0) = \frac{\sqrt{2} }{2}$

There is a maximum value of √2/2 at x = 0.

$f(\frac{7\pi}{4} ) =sin(\frac{7\pi}{4} +\frac{\pi}{4}) \\ f(\frac{7\pi}{4} ) =sin(\frac{8\pi}{4}) \\ f(\frac{7\pi}{4} ) =sin(2\pi) \\ f(\frac{7\pi}{4} ) =0$

There is a minimum value of 0 at x = 7pi/4.

We need to first compare the critical point values and then compare the endpoint values to determine whether they are maximum or minimums.

<h2>Stationary Points:</h2>

A critical point is called a stationary point if f'(x) = 0.

Since f'(x) is zero at both of the critical points, there are no critical points that are not stationary points.

6 0

3 years ago

If triangle ABC is isosceles, what is the value of ∠A?<br>

Delvig [45]

×+102=180
-102=-102
×78

4 0

3 years ago

a practice field is 250 ft long the game is 40% longer than the practice field of how long is the game ​

a practice field is 250 ft long the game is 40% longer than the practice field of how long is the game