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4 years ago

Which interval for the graphed function contains the local

Mathematics

2 answers:

Ainat [17]4 years ago

6 0

Step-by-step explanation:

The answer is [ 0 , 2 ]

But what is local maximum?

We say local maximum when there may be higher points elsewhere but not nearby.

What is local minimum?

We say local minimum when there may be lower points elsewhere but not nearby.

The general word for maximum(plural maxima) or minimum(plural minima) is extremum (plural extrema).

The maximum point is at ( 1.55 , 10.8 )

What the question means to ask in basic terms is that , between what interval of x ( on the x - axis ) can we find the maximum point if the curve .

And you will agree with me that it is between the interval of [ 0 , 2 ] , every other interval in the options doesn't contain the maximum point.

You can also visit this link to know more :

https://www.mathsisfun.com/calculus/maxima-minima.html

daser333 [38]4 years ago

4 0

Answer: The answer is [ 0 , 2 ]

Step-by-step explanation:

just did this

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3 years ago

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3 years ago

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3 years ago

There is 3/4 pound of cream in a containwr. Julian will use 3/8 pound of cream to make crème brûlée. What fraction of a pound of

Ksivusya [100]

Answer:

$\frac{3}{8}\ pound$ of cream will remain in the container.

Step-by-step explanation:

Given:

Amount of cream in the container = $\frac{3}{4}\ pound$

Amount of cream used to make crème brûlée = $\frac{3}{8}\ pound$

We need to find the amount of cream remain in the container.

Solution:

Now we can say that;

the amount of cream remain in the container is equal to Amount of cream in the container minus Amount of cream used to make crème brûlée.

framing in equation form we get;

amount of cream remain in the container = $\frac{3}{4}-\frac{3}{8}$

Now we will make the denominator common using LCM we get;

amount of cream remain in the container = $\frac{3\times2}{4\times2}-\frac{3\times1}{8\times1}= \frac{6}{8}-\frac{3}{8}$

Now denominators are common so we will solve the numerators.

amount of cream remain in the container = $\frac{6-3}{8}=\frac{3}{8}\ pound$

Hence $\frac{3}{8}\ pound$ of cream will remain in the container.

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3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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4 years ago