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3 years ago

List the sides of each triangle from shortest to greatest

Mathematics

1 answer:

True [87]3 years ago

3 0

A) First, we need to calculate x. We can do this by doing: 180-(70+50), which is nothing but 60.

Since the largest angle is 60, it means that the side opposite of it is the largest side. Which in this case, side AB.

So the largest side is AB.

Now, let's take a look at the smallest angle which is <B, because AC is the opposite of this angle, it must also be the smallest side.

We're left with angle <50, which is "medium" sized? Obviously, this can just be put in the middle (Side BC)

So, for response A, you should get AC, BC, AB.

Now for problem B, it's the same steps we did above. If you replicate what we did you should get PQ, PR,RQ.

Please let me know how I did.

Happy studying!

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A ball is thrown into the air with an upward velocity of 24 ftys. Its height h in feet after t seconds is given by the function

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And since the height is on the y-axis, the ball will each a maximum height of 16ft.

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3 years ago

Write the greatest common factor for 12 & 32

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<span>GCF (12,32) = 4
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4 years ago

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30 is what percent of 80

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3 years ago

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Ariyonne claims (3,6) that is the point of intersection of the lines y=4x-2 and y=1/2x+5 Is she correct? How do you know?

yan [13]

Answer:

The point of intersection of the lines is $(2,6)$ .

Thus, Ariyonne's claim is incorrect.

Step-by-step explanation:

Given equations:

1) $y=4x-2$

2) $y=\frac{1}{2}x+5$

To find the point of intersection, we need to solve the given system of equations.

In order to solve the system we will use substitution method.

We will substitute value of $y$ from first equation in the second equation and solve for $x$

So, we have

$4x-2=\frac{1}{2}x+5$

Multiplying both sides by 2 to remove fraction.

$2\times(4x-2)=2\times(\frac{1}{2}x+5)$

Using distribution.

$8x-4=x+10$

Subtracting both sides by $x$

$8x-4-x=x+10-x$

$7x-4=10$

Adding 4 to both sides.

$7x-4+4=10+4$

$7x=14$

Dividing both sides by 7.

$\frac{7x}{7}=\frac{14}{7}$

$x=2$

We can find value of $y$ by plugging in $x=2$ in the first equation.

$y=4(2)-2$

$y=8-2$

$y=6$

So, the point of intersection of the lines is $(2,6)$ .

Thus, Ariyonne's claim is incorrect.

3 0

3 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