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3 years ago

Determine the equation of a line that passes through the points: (3, 12) and (- 2, -13)

Mathematics

1 answer:

Lorico [155]3 years ago

7 0

Answer:

Step-by-step explanation:

(-13-12)/(-2-3)= -25/-5= 5

y - 12 = 5(x - 3)

y - 12 = 5x - 15

y = 5x - 3

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NNADVOKAT [17]

Answer:

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Step-by-step explanation:

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3 years ago

How do you substitute x+y=2 3x+2y=5

Monica [59]

Here's how you solve this. So, x+y=2, right? Let's isolate x. x+y-y=2-y. x=2-y. NOW, if x=2-y, in 3x+2y=5, we can REPLACE x with 2-y and use it to solve for y! 3x+2y=5. 3(2-y)+2y=5. (3*2)+(3*-y)+2y=5. 6+(-3y)+2y=5. 6+(-y)=5. 6+(-y)-6=5-6. -y=-1. -y/-1=-1/-1. y=1.

So, if y=1, we can substitute that back into either equation--but let's go with the easier one, x+y=2. x+1=2. x+1-1=2-1. x=1. 1+1=2, so that works; let's check the other equation. 3(1)+2(1)=5. 3+2=5. 5=5. That's correct!

Answer: x=1, y=1

4 0

3 years ago

Read 2 more answers

PLEASE ANSWER + BRAINLIEST!! Factor completely. 4k - 20k^9 = 3b^2 - 108 =

Neporo4naja [7]

$4k-20k^9=4k(1-5k^8)=4k\left[1^2-(k^4\sqrt5)^2\right]\\\\=4k(1-k^4\sqrt5)(1+k^4\sqrt5)=4k\left[1^2-(k^2\sqrt[4]5)^2\right](1+k^4\sqrt5)\\\\=4k(1-k^2\sqrt[4]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k\left[1^2-(k\sqrt[8]5)^2\right](1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k(1-k\sqrt[8]5)(1+k\sqrt[8]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)$

$3b^2-108=3(b^2-36)=3(b^2-6^2)=3(b-6)(b+6)$

$Used:\ (a-b)(a+b)=a^2-b^2$

7 0

3 years ago

6n- 15 = 6(5)? – 15 =

kozerog [31]

Answer:

what

Step-by-step explanation:

8 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}}\\

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

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\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}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{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 too easy 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.

Differential (Power Rules)

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

Differentiation (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)}}\\

\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)}}$

Differential (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}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\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}}\\

\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