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V125BC [204]
3 years ago
15

Round 562 to the nearest ten

Mathematics
2 answers:
professor190 [17]3 years ago
6 0

Answer:

560

Step-by-step explanation:

Anni [7]3 years ago
5 0

Answer: 560

Since it did not hit 565 it is 560.

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Assuming x=0 and y=0, what is the quotient of 24^6y^2 + 16x^4y^3 + 8x^2y^4
Tcecarenko [31]
There is no quotient: if x is equal to zero, then your problem would simplify to a number divided by zero. As it is impossible to divide by zero, there is no answer.

6 0
4 years ago
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The next term in the following sequence is 12. True or False<br><br>2, 4, 8…
Snezhnost [94]
False, every time in the sequence, the numbers are doubling, therefore, the next number in the sequence would be 16.
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3 years ago
Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh
Vedmedyk [2.9K]

Answer:

Increasing: x and x>1.

Decreasing: -1

Step-by-step explanation:

We have been given a function f(x)=x^3-3x+4. We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)

f'(x)=3x^{3-1}-3+0

f'(x)=3x^{2}-3

Let us equate derivative with 0 as find critical points as:

0=3x^{2}-3

3x^{2}=3

Divide both sides by 3:

x^{2}=1

Now we will take square-root of both sides as:

\sqrt{x^{2}}=\pm\sqrt{1}

x=\pm 1

x=-1,1

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9

Since 9 is greater than 0, therefore, function is increasing on interval (-\infty, -1) \text{ or } x.

Now we will check 0 for 2nd interval.

f'(0)=3(0)^{2}-3=0-3=-3

Since -3 is less than 0, therefore, function is decreasing on interval (-1,1) \text{ or } -1.

We will check 2 from our 3rd interval.

f'(2)=3(2)^{2}-3=3(4)-3=12-3=9

Since 9 is greater than 0, therefore, function is increasing on interval (1,\infty) \text{ or } x>1.

6 0
4 years ago
A sandbox is a right rectangular prism
horrorfan [7]
4.6 Is the answer I thinkkkkk
7 0
3 years ago
PLEASEEE HELLPP!!!!!!!!!!!
masya89 [10]

Answer:

D

Step-by-step explanation:

6 0
3 years ago
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