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2 years ago

Help me pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeee

Mathematics

2 answers:

hoa [83]2 years ago

8 0

Answer:

93-96 3

97-100 3

101-104 4

105-108 4

109-112 2

histogram you fill in 3 bars, 3 bars, 4 bars, 4 bars, 2 bars

Step-by-step explanation:

Aliun [14]2 years ago

6 0

Sorry can’t help I just need the points
But he will help you^

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What is the square root of 24

Rina8888 [55]

Answer:

4.9 (rounded)

Step-by-step explanation:

Square root or 24 is 4.89, so it rounds to 4.9

3 0

3 years ago

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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

What is mL <br> 38<br> 69<br> 73<br> 90

Ulleksa [173]

The size of Angle L should be the same as Angle C if the triangles are congruent (or at least similar) Since Angle C = 69o, then Angle L also = 69o. You should be very careful in accepting this assumption because looks can deceive. However if the assumption of congruency is not true, there is no way to solve the problem. L = 69o.

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3 years ago

In the lab, Debra has two solutions that contain alcohol and is mixing them with each other. Solution A is 10% alcohol and Solut

Ivahew [28]

Answer:

800 mL

Step-by-step explanation:

Let b represent the amount of solution B added to 2400 mL of solution A. The total amount of alcohol (mL) in the mixture is ...

10%×2400 + 50%×b = 20%×(2400 +b)

multiplying by 100 to get rid of the % signs and simplifying, we have ...

24000 +50b = 48000 +20b

30b = 24000 . . . . . . . . . subtract 24000+20b

b = 800 . . . . . . . . . . . . . . divide by 30

Debra added 800 mL of 50% alcohol to her 2400 mL of 10% alcohol to make a 20% solution.

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3 years ago

The quadratic function d = –3x^2 + 16 models a skateboarder’s distance, in feet, from the bottom of a hill x seconds after the s

mihalych1998 [28]

Answer:

To find the solution, plug in 8 for d

d=-3x2+16

8=-3x2+16 Subtract 8 from both sides

0=-3x2+8

Now, use the quadratic equation to solve for x

a=-3

b=0

c=8

x=1.63 or

C. 1.6s

Step-by-step explanation:

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3 years ago

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