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2 years ago

Someone please help for a brainly

Mathematics

1 answer:

ivolga24 [154]2 years ago

6 0

Answer:

it's 3 it's it

Step-by-step explanation:

trust me okay I promise dude I a easy yes yes 3

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In the figure below, m2 1=(x-12)° and mZ2 =5xº.<br> Find the angle measures.

erma4kov [3.2K]

Answer:

15 165

Step-by-step explanation:

m1 + m2 = 180

x-18 +5x = 180

6x = 198

x = 33

m1 = 15

m2 = 165

6 0

2 years ago

A fish is 4 feet under water and swims up 1 foot towards the surface. What is its position

vampirchik [111]

Answer: $\Large\boxed{-3~\text{feet under water}}$

Step-by-step explanation:

<u>Given information</u>

$\text{Fish is 4 feet under water (-4)}$

$\text{Swims up by 1 foot (+1)}$

<u>Derived expression from the given information</u>

$\text{Position = Initial + Change}$

<u>Substitute values into the given expression</u>

$\text{Position = }(-4)+(1)$

<u>Simplify the position</u>

$\text{Position = }-3$

<u>Therefore, the current position of the fish is:</u>

$\Large\boxed{-3~\text{feet under water}}$

Hope this helps!! :)

Please let me know if you have any questions

7 0

1 year ago

Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7

lilavasa [31]

Answer:

<h2>A.Vertical:x=7</h2><h2>Slant:y=x+9</h2>

Step-by-step explanation:

$f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}$

$slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1$

$b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}$