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3 years ago

7 times 10 to the -7 power

Mathematics

1 answer:

klasskru [66]3 years ago

3 0

Answer:

0.0000007 as a regular notation

or 7x10^-7 as scientific notation

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2 years ago

Describe the difference between total surface area and lateral surface area

Diano4ka-milaya [45]

Answer:

The total surface area of a solid is the sum of the areas of all of the faces or surfaces that enclose the solid, the lateral surface area of a solid is the surface area of the solid without the bases

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3 years ago

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natima [27]

Answer:

what are yo tring to sya

Step-by-step explanation:

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3 years ago

Select a setting so that each given point will lie within the viewing window.

NemiM [27]

Answer:

The given points are

$(-13,3)\\(0,0)\\(-2,-7)$

The setting would have a interval or 2 units above and below the minimum and maximum of each coordinate.

The given maxium horizontal coordinate is 0.

The given minimum horizontal coordinate is -13.

The given maximum vertical coordinate is 3.

The given minimum vertical coordinate is -7.

Now, we extend each maximum and minimum value by 2 units to create the setting.

So, the setting is

$x_{min}=-15\\x_{max}=2\\ y_{min}=-9\\y_{max}=5$

With a scale of 2 units.

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3 years ago

*Asymptotes* g(x) =2x+1/x-3 Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

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3 years ago