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3 years ago

Approximate 650.385 to 1 decimal place

Mathematics

2 answers:

marishachu [46]3 years ago

8 0

Rounded to one decimal place would be 650.4

stealth61 [152]3 years ago

7 0

650.385 rounded to one decimal place is 650.4

brainiest?

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Solve. Good luck! Please do not try to google this.

Elena L [17]

$\frac{x^{2} + x - 2}{6x^{2} - 3x} = \sqrt{2x} + \frac{3x^{2}}{2}$
$\frac{x^{2} + 2x - x - 2}{3x(x) - 3x(1)} = \frac{2\sqrt{2x}}{2} + \frac{3x^{2}}{2}$
$\frac{x(x) + x(2) - 1(x) - 1(2)}{3x(x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{x(x + 2) - 1(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$3x(2x - 1)(2\sqrt{2x} + 3x^{2}) = 2(x + 2)(x - 1)$
$3x(2x(2\sqrt{2x} + 3x^{2}) - 1(2\sqrt{2x} + 3x^{2}) = 2(x(x - 1) + 2(x - 1))$
$3x(2x(2\sqrt{2x}) + 2x(3x^{2}) - 1(2\sqrt{2x}) - 1(3x^{2})) = 2(x(x) - x(1) + 2(x) - 2(1)$
$3x(4x\sqrt{2x} + 6x^{3} - 2\sqrt{2x} - 3x^{2}) = 2(x^{2} - x + 2x - 2)$
$3x(4x\sqrt{2x} - 2\sqrt{2x} + 6x^{3} - 3x^{2}) = 2(x^{2} + x - 2)$
$3x(4x\sqrt{2x}) - 3x(2\sqrt{2x}) + 3x(6x^{3}) - 3x(3x^{2}) = 2(x^{2}) + 2(x) - 2(2)$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} + 18x^{4} - 9x^{3} = 2x^{2} + 2x - 4$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} = -18x^{4} + 9x^{3} + 2x^{2} + 2x - 4$
$6x\sqrt{2x}(2x) - 6x\sqrt{2x}(1) = -9x^{3}(2x) - 9x^{3}(-1) + 2(x^{2}) + 2(x) - 2(2)$
$6x\sqrt{2x}(2x - 1) = -9x^{3}(2x - 1) + 2(x^{2} + x - 2)$

5 0

4 years ago

HELP!!!!!!!!!!!!!!!!!!

padilas [110]

First, factor each expression by finding two terms whose product is equal to the c term and sum is equal to the b term.

x^2-x+6 --> (x-3)(x+2)

x^2-4 --> (x+2) (x-2)

Next, cancel the terms that are equal on the top and bottom.

You are left with the answer of: $\frac{x-3}{x-2}$

Hope this helps!!

8 0

3 years ago

Consider the sequence: 4, 7, 10, ... , ... , ....<br><br> What is the nth term of this sequence?

Tju [1.3M]

4+3=7+3=10+3=13. Steps.........

Ninth term in the sequence is 30

4 0

4 years ago

Read 2 more answers

Simplify"<br>2x² + 16x + 14 / X² - 1

Phoenix [80]

Answer: 16x+3+x214

Step-by-step explanation:

Multiply 2 and 2 to get 4.

4+16x+x214−1

Subtract 1 from 4 to get 3.

3+16x+x214

To add or subtract expressions, expand them to make their denominators the same. Multiply 3+16x times x2x2.

x2(3+16x)x2+x214

Since x2(3+16x)x2 and x214 have the same denominator, add them by adding their numerators.

x2(3+16x)x2+14

Do the multiplications in (3+16x)x2+14.

x23x2+16x3+14

5 0

3 years ago

Select all of the transformations that could change the location of the asymptotes of a cosecant of secant function.

mars1129 [50]

We have to select all of the transformations that could change the location of the asymptotes of a cosecant of secant function.

So given function can be written as:

y=csc( sec(x))

First we need to determine the location of asymptote which is basically a line that seems to be touching the graph of function at infinity.

From attached graph we see that Asymptotes (Green lines) are vertical.

So Vertical shift or vertical stretch will not affect the location of asymptote because moving up or down the vertical line will not change the position of any vertical line.

only Left or right side movement will change the position of vertical asymptote. Which is possible in Phase shift and period change.

Hence Phase shift and Period change are the correct choices.

8 0

4 years ago

Read 2 more answers