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dimulka [17.4K]
4 years ago
7

Roumd 33 to the nearest 100​

Mathematics
1 answer:
storchak [24]4 years ago
6 0

Answer:

Step-by-step explanation:

Trick question. Good to know.

0 is the closest 100.

33 will round to 0

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1 1/4 +(3 2/3 +5 3/4)​
andrew11 [14]
10 2/3. if you need it in fraction form it's 32/3, and if you need it in decimal form it's 10.6.
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What is the volume of the block if each side measures 1 foot?
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The volume for a cube is the cube of any side. side^3

If each sides measures 1 foot, the volume would be 1 ft^3.
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(-7) • (-5) what does it equal to
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Answer:

35

Step-by-step explanation:

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What is the monomial if a square of a monomial is: c 0.0001a^8x^4
konstantin123 [22]

Answer:

The answer to your question is 0.1a⁴x²

Step-by-step explanation:

                                           0.0001a⁸x⁴

Process

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2.- Get the square root of a⁸

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3.- Get the square root of x⁴

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4.- Write the solution

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3 0
4 years ago
Here we will study the function f (x) = e ^ x sin (x), where x ∈ [0, 2π]. a) Determine where the function is decreasing and incr
Semmy [17]

fff

f(x) = e^x sin (x)

To find increasing and decreasing intervals we take derivative

f'(x) = e^xsin(x)+e^x(cosx)= e^x(sinx+cosx)

Now we set the derivative =0  and solve for x

e^x(sinx+cosx)=0

sinx + cosx =0

divide whole equation by cos x

\frac{sinx}{cosx} + \frac{cosx}{cosx} =0

tanx +1 =0

tanx = 1

x=\frac{3\pi }{4} and  x=\frac{7\pi}{4}

Now we pick a number between 0 to  \frac{3\pi }{4}

Lets pick  \frac{\pi }{2}

Plug it into the derivative

f'(x) =e^{\frac{\pi }{2}}(sin(\frac{\pi}{2})+cos(\frac{\pi }{2}))

= 4.810 is positive

So the graph of f(x) is increasing on the interval [0, x=\frac{3\pi }{4})

Now we pick a number between   \frac{7\pi}{4} to 2pi

Lets pick  \frac{11\pi}{6}

Plug it into the derivative

f'(x) =e^{\frac{11\pi}{6}}(sin(\frac{11\pi}{6})+cos(\frac{11\pi }{6}))

= 116 is positive

So the graph of f(x) is increasing on the interval (\frac{7\pi }{4}, 2\pi)

Increasing interval is (0,\frac{3\pi }{4}) U (\frac{7\pi }{4}, 2\pi)

Decreasing interval is (\frac{3\pi}{4}, \frac{7\pi}{4})

(b)

The graph of f(x) increases and reaches a local maximum at x=\frac{3\pi}{4}

The graph of f(x) decreases and reaches a local minimum at x=\frac{7\pi}{4}

(c)

f(0) = 0

f(2\pi)=0

f(\frac{3\pi }{4})=7.46

f(\frac{7\pi}{4})=-172.64

Here global maximum at x=\frac{3\pi}{4}

Here global minimum at x=\frac{7\pi}{4}


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3 years ago
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