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2 years ago

What is the value of 0.1322 rounded to the nearest hundredth place?

Mathematics

2 answers:

frosja888 [35]2 years ago

8 0

Answer:

0.13

Step-by-step explanation:

First we need to find the given place value, which is the hundreds place.

0.1322

Next, we look at the next number over, which is two.

0.1322

0-4 stays the same. 5-9 round up one. Doing so, we keep the number three the same.

0.1300

get rid of the extras.

(not a Bakogou reference)

0.13

~Hope this helps~

kifflom [539]2 years ago

3 0

Answer:

Step-by-step explanation:

0.1322 ≅ 0.13

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Consider the function below. (If an answer does not exist, enter DNE.) g(x) = 190 + 8x3 + x4 (a) Find the interval of increase.

galben [10]

Answer:

The interval of increase of g(x) is $(-6,+\infty)$ .

Step-by-step explanation:

The interval of increase occurs when first derivative of given function brings positive values. Let be $g(x) = 190 + 8 \cdot x^{3} + x^{4}$ , the first derivative of the function is:

$g'(x) = 24 \cdot x ^{2} + 4\cdot x^{3}$

$g'(x) = 4 \cdot x^{2}\cdot (6+x)$

The following condition must be met to define the interval of increase:

$4\cdot x^{2} \cdot (x+6) > 0$

The first term is always position due to the quadratic form, the second one is a first order polynomial and it is known that positive value is a product of two positive or negative values. Then, the second form must satisfy this:

$x + 6 > 0$