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mafiozo [28]
2 years ago
7

What is the exact distance from (-5,1) to (3,0).

Mathematics
1 answer:
Ilya [14]2 years ago
5 0

Answer:

The distance between the two points is \sqrt{65} \ \text{units}.

Step-by-step explanation:

In order to find the distance between two coordinate pairs, we can use the distance formula:

\displaystyle \bullet  \ \ \ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Our coordinate pairs need to be labeled accordingly, so we can use this naming system:

\bullet  \ \ \ (x_1, y_1), (x_2, y_2)

This assigns a name to our points:

  • x_1 = -5
  • y_1 = 1
  • x_2 = 3
  • y_2 = 0

Therefore, we can plug these into the formula and solve:

d=\sqrt{(3-(-5))^2+(0-1)^2}\\\\d=\sqrt{(8)^2+(-1)^2}\\\\d=\sqrt{8^2+1^2}\\\\d=\sqrt{64+1}\\\\d=\sqrt{65}

Therefore, the distance between the two points is \sqrt{65} \ \text{units}.

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3 years ago
Determine the asymptotes of the function: y=x^3-5x^2+4x-25/x^2-4x+3
igomit [66]

Answer:

Vertical A @ x=3 and x=1

Horizontal A nowhere since degree on top is higher than degree on bottom

Slant A @ y=x-1  

Step-by-step explanation:

I'm going to look for vertical first:

I'm going to factor the bottom first:  (x-3)(x-1)

So we have possible vertical asymptotes at x=3 and at x=1

To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)

3^3-5(3)^2+4(3)-25=-31 so it isn't a factor of the top so you have a vertical asymptote at x=3

Let's check x=1

1^3-5(1)^2+4(1)-25=-25 so we have a vertical asymptote at x=1 also

There is no horizontal asymptote because degree of top is bigger than degree of bottom

There is a slant asympote because the degree of top is one more than degree of bottom (We can find this by doing long division)

                       x   -1

                --------------------------------------------------

x^2-4x+3 |      x^3-5x^2+4x-25

                  - ( x^3-4x^2+3x)

                   --------------------------------

                            -x^2 +x  -25

                       -   (-x^2+4x-3)

                          ---------------------

                                   -3x-22

So the slant asymptote is to x-1

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3 years ago
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Find the domain of the function f(x)=Log(|x-7|-3) and give the solution in interval notation
Alecsey [184]

Answer:

(-∞, 4) ∪ (10, ∞)

Step-by-step explanation:

A logarithm y = log x can only take x-values greater than 0. It's domain is (0, ∞). If you put log(0) or log of any negative number into a calculator, it'll return an error. So when you have a logarithm like yours, f(x) = log(|x - 7| - 3), you need to make sure you take out any x value that could give you a negative result or 0.

Take the argument of the logarithm, |x - 7| - 3, and pull it aside. You can solve it one of two ways: by looking at it and plugging in values and figuring out which values give you zero or a negative number (trial and error), or by setting it equal to zero and solving for x. I'll show you the second approach:

|x - 7| - 3 = 0 ... get the absolute value bars alone

|x - 7| = 3 ... absolute value means "this distance from 0 on a number line", so what your equation means is |x - 7| is 3 units away from zero. But we don't know if it's three units to the left (negative 3) or three units to the right (positive three). So we set up both.

x - 7 = 3

x - 7 = -3

Solve both equations, and get x = 10 for the top equation, and x = 4 for the bottom equation. This gives you the values you need to use in your interval notation. 4 and 10 both will give you 0 if you plug them into |x - 7| - 3, and 0 is outside of the domain of a logarithm. You can't have negative numbers either, and you'll see that if you try to plug in 5, 6, 7...all the way up to 10, you'll either get a negative result or 0. Basically, anything between 4 and 10, including 4 and 10 won't do it for you. x has to be less than 4 or greater than 10 for your original f(x) function to be defined.

Now that you've identified what ISN'T in your domain, all you have to do is set it up in interval notation. You're clear for everything from negative infinity up to 4, but at exactly 4 (if you set x = 4 and plug it in), your result is undefined. Exactly at 10, it's undefined, but anything greater than 10 works fine. So (-∞, 4) ∪ (10, ∞) is your answer.

8 0
3 years ago
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