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3 years ago

In a herd of cattle, the ratio of the number

Mathematics

1 answer:

Neko [114]3 years ago

7 0

Answer:

In a herd no. of Cattle are: x

The ratio of bulls to cows is 1:6

\frac{bulls}{cows} = \frac{1}{6}

6 bulls = 1 cows

bulls \: + cows = x

bulls + 6 \: bulls \: = x

7bulls = x

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Select all the situations that can be represented by 1/2⋅5. * 1 point Diego lives 5 miles from school, and Elena lives 1/2 as fa

SpyIntel [72]

Answer:

Elena lives 2.5 miles away from school.
The total liters of water is 2.5 liters
He can cut 10 pieces of rope of length 1/2 meter
The fraction of her goal she ran is 1/10

Step-by-step explanation:

From the question,

Diego lives 5 miles from school, and Elena lives 1/2 as far away as Diego

Therefore, Elena lives 1/2 of 5 miles away from school.

1/2 of 5 miles = 2.5 miles

Hence, Elena lives 2.5 miles away from school.

Jada has 5 bottles that each contain 1/2 liter of water.

Since each bottle contain 1/2 liter of water, then

The total liters of water is (1/2 × 5) liters

1/2 × 5 = 2.5 liters

Hence, the total liters of water is 2.5 liters

Noah has 5 meters of rope

To determine how many pieces of rope of length 1/2 meter he can cut,

Number of pieces of rope of length 1/2 meter that can be cut is 5 meter ÷ 1/2 meter

5 meter ÷ 1/2 meter = 10

Hence, he can cut 10 pieces of rope of length 1/2 meter.

Lin's goal is to run 5 miles. She ran 1/2 mile.

To determine what fraction of her goal that is,

First, she ran 1/2 mile out of 5 miles.

The fraction of her goal she ran is 1/2 mile ÷ 5 miles

1/2 mile ÷ 5 miles = 1/10

Hence, the fraction of her goal she ran is 1/10.

8 0

4 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

the sum of the measurement of two angles is 183.61 angle one has a measure of 54 what is the measure in degrees of the second an

Dennis_Churaev [7]

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7 0

3 years ago

Please help fraction form read

pogonyaev

The answer to the question is 4/15

8 0

3 years ago

Read 2 more answers

Devante buys 1/1/4 pounds of cashews 7 ounces of almonds and 3/4 pounds of pecans how much do the nuts weight in

liberstina [14]

Answer: $39\ \text{Ounces}$

Step-by-step explanation:

Given

The amount of cashews is $1\frac{1}{4}\ \text{Pounds}$

Amount of almonds $7\ \text{ounces}$

The amount of pecans is $\frac{3}{4}\ \text{Pounds}$

We know, $1\ \text{Pound}=16\ \text{Ounces}$

Converting pounds to ounces and adding them

$\Rightarrow 1\dfrac{1}{4}\times 16+7+\dfrac{3}{4}\times 16\\\\\Rightarrow \dfrac{5}{4}\times 16+7+3\times 4\\\\\Rightarrow 20+7+12=39\ \text{ounces}\ or\ \dfrac{39}{16}\ \text{Pounds}$

Thus, the nuts weigh $39\ \text{Ounces}$

6 0

3 years ago