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2 years ago

Classify the following triangle as acute, obtuse, or right

Mathematics

1 answer:

kherson [118]2 years ago

8 0

Answer:

Acute

Step-by-step explanation:

It should be acute as its less than 90 (right angle) and that also mean its not obtuse ( any angle more than 90.

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PLEASE HELP ME!!! PLEASE TRY TO DO ALL OF THEM THX! :)

kakasveta [241]

Hello there!

Number 1 is C (11, 4) Image is below

I don't know number 2, so so sorry!

Number 3 is A. 36 The formula for perimeter of a rectangle is (2 * base) + (2 * height)
The base = 13
The height = 5

(5 * 2) + (13 * 2)

10 + 26 = 36

Number 4 is 23 units.

I hope I was of assistance!

~ Fire

5 0

3 years ago

The sketch shows two parallel lines cut by a transversal. Which set of angle pairs are NOT corresponding angles?

gayaneshka [121]

They both aren't. because no right angles.

8 0

3 years ago

Read 2 more answers

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:

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

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

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

$=\lim\limits_{x\to\pm\infty}\dfrac{9-\frac{3}{x}}{1-\frac{7}{x}}=\dfrac{9}{1}=9\\\\\boxed{y=1x+9}$

8 0

3 years ago

What is 5.3809 rounded to the nearest thousandth?<br>Enter your answer

Hatshy [7]

Answer:

5.381

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

An exterior angle of an isosceles triangle has measure 130°. Find two possible sets of measures for the angles of the triangle.

frozen [14]

(1) it can be both exterior of vertex or base. 180-130=50°
vertex is 50°: base=(180-50)/2=65°
base is 50°: vertex= 180-2*50=80°

vertex 50 base 65; vertex 80, base 50

(2)base=180-130=50°
vertex=180-2*50=80°

base 50 vertex 80

5 0

3 years ago