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3 years ago

12

0.96 in fraction form

1 answer:

sineoko [7]3 years ago

8 0

96/100, Is .96 in fraction form.
I hope this helps

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How can you use division to find the decimal equivalent of a rational number ?

Naily [24]

Answer:

Rational numbers' decimal representations either finish or repeat a pattern. Divide the top by the bottom, the numerator by the denominator to convert fractions to decimals, and if the division doesn't come out evenly, round off after a specific number of decimal places.

Step-by-step explanation:

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2 years ago

Write numbers to make each line have the same sum

wlad13 [49]

Answer:

There is nothing here?

Step-by-step explanation:

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2 years ago

Is 2y-x/6-4=0 a linear function

noname [10]

A equation of a linear function in a standard form:

Ax + By = C

$2y-\dfrac{x}{6}-4=0\ \ \ \ |\cdot(-6)\\\\-12y+x+24=0\ \ \ \ |-24\\\\-12y+x=-24\\\\\boxed{x-12y=-24}$

<h3>Answer: YES. It's a linear function.</h3>

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3 years ago

2x+2y=8<br> X+y=4<br> Which statement is true about the solution

laila [671]

Both x and y equal 2.

2•2=4 + 2•2=4

4+4=8

2+2=4

6 0

3 years ago

please show on graph (with x and y coordinates) state where the function x^4-36x^2 is non-negative, increasing, concave up

babunello [35]

Answer:

$y'' =12x^2 -72=0$

And solving we got:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

We can find the sings of the second derivate on the following intervals:

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up

Step-by-step explanation:

For this case we have the following function:

$y= x^4 -36x^2$

We can find the first derivate and we got:

$y' = 4x^3 -72x$

In order to find the concavity we can find the second derivate and we got:

$y'' = 12x^2 -72$

We can set up this derivate equal to 0 and we got:

$y'' =12x^2 -72=0$

And solving we got:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

We can find the sings of the second derivate on the following intervals:

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up

8 0

2 years ago