If the radius of mars is about 13.7% of neptune's radius, what is the radius of neptune?

Answers

Answer 1

Answer:
You said that M is 13.7% of N

In algebraic notation M = 0.137 N

Divide each side by 0.137 : N = M / 0.137 = 7.3 M (rounded)

Online or in a real book, you can look up the radius of Mars.
Then write it in place of 'M' in this solution, and you'll have
an actual number for the radius of Neptune.

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Which phrase best defines a square? A.
a quadrilateral with exactly one pair of parallel sides

B.
a quadrilateral with no congruent sides

C.
a rectangle with four congruent sides

D.
a parallelogram with four congruent angles
I think its D?

Answers

the answer is D a parallelogram with four congruent angles

Find the roots of the polynomial equation -x^3+5x^2 - 11x+55=0

Answers

  x³ + 5x² - 11x + 55 = 0
x²(x) + x²(5) - 11(x) - 11(5) = 0
  (x²(x + 5) - 11(x + 5) = 0
(x² - 11)(x + 5) = 0
x² - 11 = 0     or        x + 5 = 0
+ 11 + 11 - 5 - 5
x² = 11 or x = -5
  x = ±√(11) or x = -5

THere are three roots in the polynomial equation.

What is the midpoint of FB

Answers

Answer:

3. 15. . 16..

If this answer helps you plz mark as brainlist..

Answer:

D or L is the midpoint

Step-by-step explanation:

What is the volume of 5 m 8m and 10.5m

Answers

Answer:

Volume = 5×8×10.5 = 420 meters3

Step-by-step explanation:

5×8×10.5

Help with line equations?

Answers

First question

$line:~~~~y=x+3$

$curve:~~~~x^2+y^2=29$

now we have to replace the y of curve by the y of line, therefore

$x^2+(x+3)^2=29$

$x^2+x^2+6x+9=29$

$2x^2+6x+9-29=0$

$2x^2+6x-20=0$

we can multiply each member by $(1)/(2)$

$\boxed{\boxed{x^2+3x-10=0}}$

Now we have to find the roots of this funtion

$x^2+3x-10=0$

Sum and produc or Bhaskara

Then we find two axis

$\boxed{x_1=2~~and~~x_2=-5}$

now we replace this values to find the Y, we can replace in curve or line equation, I'll prefer to replace it in line equation.

$y=x+3$

$y_1=x_1+3$

$y_1=2+3$

$\boxed{y_1=5}$

$y_2=x_2+3$

$y_2=-5+3$

$\boxed{y_2=-2}$

Therefore

$\boxed{\boxed{P_1(2,5)~~and~~P_2(-5,-2)}}$
_______________________________________________________________

The second question give to us

$y=ax+b$

$P_1(2,13)$

$P_2(-1,-11)$

We just have to replace the value then we'll get a linear system.

point 1

$13=2a+b$

point 2

$-11=-a+b$

then our linear system will be

$\begin{Bmatrix}2a+b&=&13\n-a+b&=&-11\end{matrix}$

I'll multiply the second line by -1 and I'll add to first one

$\begin{Bmatrix}2a+(a)+b+(-b)&=&13+(-11)\n-a+b&=&-11\end{matrix}$

$\begin{Bmatrix}3a&=&24\n-a+b&=&-11\end{matrix}$

$\begin{Bmatrix}a&=&8\n-a+b&=&-11\end{matrix}$

therefore we can replace the value of a, at second line

$\begin{Bmatrix}a&=&8\n-8+b&=&-11\end{matrix}$

$\begin{Bmatrix}a&=&8\nb&=&-11+8\end{matrix}$

$\boxed{\boxed{\begin{Bmatrix}a&=&8\nb&=&-3\end{matrix}}}$

then our function will be

$\boxed{\boxed{y=8x-3}}$

_________________________________________________________________

The third one, we have

$line:~~~~y=3x-4$

$curve:~~~~y=x^2-2x-4$

This resolution will be the same of our first question.

Let's replace the y of curve by the y of line

$3x-4=x^2-2x-4$

$0=x^2-2x-4-3x+4$

therefore

$\boxed{\boxed{x^2-5x=0}}$

now we have to find the roots of this function.

$x^2-5x=0$

put x in evidence

$x*(x-5)=0$

$\boxed{x_1=0~~and~~x_2=5}$

then

$y=3x-4$

$y_1=3x_1-4$

$y_1=3*0-4$

$\boxed{y_1=-4}$

$y_2=3x_2-4$

$y_2=3*5-4$

$y_2=15-4$

$\boxed{y_2=11}$

our points will be

$\boxed{\boxed{P_1(0,-4)~~and~~P_2(5,11)}}$

I hope you enjoy it ;)

What is the slope of a line that is perpendicular to the line y = -x + 4?The slope of the line is
y=5-6x

Answers

Answer:

1 is the slope

Step-by-step explanation:

For perpendicular lines, you need to find the negative recipricol which would be 1 for -1

Answer:

y=x+4

Step-by-step explanation:

just flipped the sign of the slope and the line will go the other way

this question is worded a little weird because the y is not what represents slope in the formula so. Assuming that, the answer to the first part should be correct

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