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Anarel [89]
3 years ago
9

The distance formula, , is the key element in the proof of which of the following?

Mathematics
1 answer:
enyata [817]3 years ago
8 0
I would say B.

Hope this helped! 

:)
You might be interested in
How do you solve 7.86 x 4.6
mihalych1998 [28]
<span><u><em>Answer:</em></u>
36.156<span>

<u><em>Explanation:</em></u>
Multiply as 7.86 and 4.6 ignoring the decimals for now. You get a solution of 36156.
You count how many decimal points are in the problem (2 from 7.86 and 1 from 4.6) and you put it in the answer by moving the decimal to the left the amount of decimals in the problem this leave the answer as 36.156</span></span>
7 0
3 years ago
Please help. 24 points!!
Lina20 [59]

The given function is,

f(x)= \frac{2x-1}{x+5}.

Or, y= \frac{2x-1}{x+5}

To find the inverse of a function, first step is to switch x and y. Therefore,

x= \frac{2y-1}{y+5}

Next stpe is to solve the above equation for y t get the inverse f f(x). Hence, multiply each sides by y + 5 to get rid of fraction form. So,

x*(y+5)= \frac{2y-1}{y+5}*(y+5)

xy + 5x = 2y - 1

xy = 2y - 1 - 5x Subtract 5x from each sides to isolate y.

xy - 2y = -5x - 1 Subtract 2y from each sides.

y(x - 2) = -5x -1 Take out the common factor y.

\frac{y(x-2)}{(x-2)} =\frac{(-5x-1)}{(x-2)} Divide each sides by x-2.

y=\frac{(-5x-1)}{(x-2)}

So, f^-1(x)=\frac{-5x-1}{x-2}

Now when we will compare f^-1(x)=\frac{-5x-1}{x-2} with f^-1(x)=\frac{ax+b}{cx+d} then we will get a = -5, b = -1, c = 1 and d=-2

So, \frac{a}{c} =\frac{-5}{1} =-5

Hope this helps you!

6 0
3 years ago
Robert drives 45 miles in 3 hours from New York City to New Jersey. How far will he drive in 5 hours?
fenix001 [56]
Ok so the awnser is 75 because 45 ÷ 3 is 15 and so 2 more hours so I added 30 to the 45 and I got 75 I think that's right please correct me
3 0
3 years ago
Need help on this ASAP
serg [7]

Answer:

so what to fo tell in brief

if u have time

3 0
3 years ago
Find two unit vectors orthogonal to both given vectors. i j k, 4i k
Maurinko [17]
The cross product of two vectors gives a third vector \mathbf v that is orthogonal to the first two.

\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k

Normalize this vector by dividing it by its norm:

\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)

To get another vector orthogonal to the first two, you can just change the sign and use -\mathbf v.
6 0
3 years ago
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