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Bas_tet [7]
3 years ago
13

The perpendicular bisector of a line segment is a line perpendicular to the segment that passes through its midpoint. What is th

e equation for the perpendicular bisector of segment XY with point X(2, -6) and Y(5, -5)?
Mathematics
1 answer:
adoni [48]3 years ago
8 0

Answer:

y = 1/3 x.

Step-by-step explanation:

The slope of the given line is

(4 - (-2) / 2 - 4)

=   -3.

So the slope of the line perpendicular to it = -1 / (-3) = 1/3.

This line also passes through the point (3, 1) so its equation is ( by the point-slope form):

y - 1 = 1/3(x - 3)

y - 1 = 1/3x - 1

y = 1/3 x (answer).

hope that help you mark me brinilylist

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Which of the following, is not a solution for the inequalities is represented by the graph shown?
stealth61 [152]

Answer:

(0,1)

Step-by-step explanation:

This is because (0,1) is right on the line, while the other choices are inside of the blue shaded area.

If this answer is correct, please make me Brainliest!

3 0
3 years ago
What is the answer to this???
Katarina [22]
A = √(15²-9²) = √(225-81) = √144 = 12
5 0
3 years ago
Read 2 more answers
Is this proportion true 12/15=47/50<br>16/36=12/27<br>15/20=24/26<br>or 4/16=2/14
jekas [21]
For <span>12/15=47/50, we multiply 12/15 by (1/3)/(1/3) to get 4/5. For 47/50, we multiply it by (1/10)/(1/10) to get the same denominator and 4/7/5. They are not equal.

For 16/36 and 12/27, we'll notice that 36 and 27 are both multiples of 9, so we need to get that as the denominator! Multiply 16/36 by (1/4)/(1/4) because 9 goes into 36 4 times and 12/27 by (1/3)/(1/3) due to that 27 goes into 9 3 times, we get 4/9=4/9 - these are equal!

I challenge you to get the rest of them on your own using my techniques!</span>
3 0
3 years ago
A teacher wrote this experssion on the board (-6)(2)+(-8÷4) what is the vaule of this expression
VLD [36.1K]

Answer:

-14

Step-by-step explanation:

(-12)+(-2)

You solve using Pemdas, doing the (-8/4) first, and then doing (-6)(2)

The answer is the two values added together.

3 0
2 years ago
How do you find the limit?
coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}

We can cancel x-5.

\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}

Then directly substitute x = 5 in.

\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}

<u>Differentiation</u> (Property of Addition/Subtraction)

\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}

Hence from the expressions,

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}

<u>Differential</u> (Constant)

\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \  is \ \ a \ \ constant.)}}

Therefore,

\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}

Now we can substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}

Thus, the limit value is 2/5 same as the first method.

Notes:

  • If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.
8 0
3 years ago
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