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3 years ago

Help...?! Pleassssssssssssse

Mathematics

1 answer:

sveta [45]3 years ago

4 0

1. -10 & 4
2. 20 days
4. 12 minutes
7. 25
10. -
14. > (greater than)

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Can anyone help me with this please?

USPshnik [31]

Answer:

Step-by-step explanation:

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4 years ago

Please help ASAP!!!!

eimsori [14]

Answer:

It'll have an open circle on both 7 and 21, with a line connecting.

Which is none of the options, unless one isn't shown?

Step-by-step explanation:

-23 > -4x + 5 > -79

So:

-23 > -4x + 5

-23 - 5 > -4x

-4x < -28

x > -28 / -4

x > 7

-4x + 5 > -79

-4x > - 79 - 5

-4x > -84

x < -84 / -4

x < 21

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3 years ago

Identify the range of the functions shown in the graph

miss Akunina [59]

Answer:

A. y >= 0

Step-by-step explanation:

The range of a function is the set of all values that the y-coordinate can have.

Look at the graph. Look at the red curve. The lowest y-coordinate is 0. As the curve goes up to the right, the y-coordinates are all real numbers greater than or equal to 0.

Answer: A. y >= 0

7 0

4 years ago

Read 2 more answers

What is Q3 for this data set?

Kisachek [45]

The same as last time

Step-by-step explanation:

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2 years ago

Read 2 more answers

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

3 0

3 years ago