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3 years ago

12

Neptune’s average distance from the sun is 4.503 x 10e9 km. Mercury’s average distance from the sun is 5.791 x10e9 km.about how

many times farther from the sun is Neptune than mercury?

1 answer:

Igoryamba3 years ago

4 0

Answer:

77.76 times

Step-by-step explanation:

The average distance of Neptune from the sun

= 4.503 × 10 ⁹ k m .

and Mercury = 5.791 × 10 ⁷ k m .

Hence neptune is ( 4.503 × 10 ⁹) ÷ (5.791 × 10 ⁷ ) times farther from the sun than mercury

i.e.( $\frac{4.503}{5.791}$ ) × 10⁹⁻⁷ times

= 0.7776 × 10 ² times

= 77.76 times.

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determine a formula for the exponential function whose graph passes through the points (0,3)and (2,6)

Black_prince [1.1K]

The initial value is 3 (when x=0), and the multiplier is 2, when x=2. The equation can be written as
.. y = 3*(2^(x/2))

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2 years ago

When making a book cover, Anwar adds an additional 20 square inches of paper will Anwar used to make a cover for a book 11 inche

Rudik [331]

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3 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> 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.)

6 0

3 years ago

Write an equation for the circle with endpoints of a diameter (9, 4) and (-3, -2)

inysia [295]

If the diameter is at (9,4) and (-3,-2), then the center of the circle is the midpoint of that segment.

and since we know that the radius of a circle is half of the diameter, whatever long that diameter segment is, the radius is half that.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 9 &,& 4~) 
% (c,d)
&&(~ -3 &,& -2~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-3+9}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \stackrel{center}{(3~,~1)}$

$\bf -------------------------------\\\\
~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 9 &,& 4~) 
% (c,d)
&&(~ -3 &,& -2~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2}
\\\\\\
d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}$

$\bf -------------------------------\\\\
\textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{3}{ h},\stackrel{1}{ k})\qquad \qquad 
radius=\stackrel{\frac{\sqrt{180}}{2}}{ r}
\\\\\\
(x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2}
\\\\\\
(x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45$

7 0

3 years ago

What is the equation of the line that passes through the point (2, -2) and has a<br> slope of -1/2?

yanalaym [24]

Answer:

Step-by-step explanation:

use the point-slope formula [y-y1 =m(x-x1) ] with the given info

for the point P=(x1,y1) =(2,-2) and the slople m = -1/2

plug those into the formula

y-(-2)= -1/2(x-2)

y+2= -1/2x + 1

y= -1/2x+1 - 2

y = -1/2x - 1

:) got it?

these are hard untill you get the idea of the two formulas

point-slope: y-y1 = m(x-x1) and

slope-intercept: y= mx+b

memorizing those will help a lot

3 0

2 years ago