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3 years ago

Find the area of this shape

Mathematics

1 answer:

nadya68 [22]3 years ago

8 0

Answer:

Step-by-step explanation:

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15 points!!! GEOMETRY help please!! Questions attached!

dolphi86 [110]

3. Look at the picture.

We have the right angle triangle. We know the sum of measures of angles in triangle is equal 180°. Therefore:

$x^o+90^o+43^o=180^o\\\\x^o+133^o=180^o\ \ \ |-133^o\\\\x^o=47^o$

$Answer:\ \boxed{47^o}$

4.
Look at the picture.

Use Pythagorean theorem:

$x^2+14^2=(10+x)^2\\\\x^2+196=10^2+2\cdot10\cdot x+x^2\ \ \ \ |-x^2\\\\196=100+20x\ \ \ |-100\\\\20x=96\ \ \ |:20\\\\x=4.8$

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

$Answer:\ \boxed{4.8\ units}$

5.
TRUE: 1; 2; 4

6.
We find a slope of the line OP:

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\O(2;\ 6)\to x_1=2;\ y_1=6\\\\P(4;\ 3)\to x_2=4;\ y_2=3\\\\m=\dfrac{3-6}{4-2}=\dfrac{-3}{2}$

We have: $OP:\ y=-\dfrac{3}{2}x+b$

Now, we must find the slope of the line perpendicular to the line OP.
We know:

$k:y=m_1x+b;\ l:y=m_2x+c\\\\k\ \perp\ l\iff m_1m_2=-1$

therefore

$-\dfrac{3}{2}m_2=-1\ \ \ |\cdot\left(-\dfrac{2}{3}\right)\\\\m_2=\dfrac{2}{3}$

So. We have the answer! :)

$Answer:\ \boxed{y=\dfrac{2}{3}x+\dfrac{1}{3}}$

6 0

3 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$