How are piecewise functions and absolute value functions related?

The three angles of a triangle create the extended ratio 2:2:5. What is the measure of the largest angle?

Dima020 [189]

Answer:

100°

Step-by-step explanation:

The ratio of angles of the triangle is 2:2:5. We know that the angle sum property of the triangle is 180°. So that , let the ratio be 2x : 2x : 5x .

=> 2x + 2x + 5x = 180°

=> 9x = 180°

=> x = 180°/9

=> x = 20°

Here the largest ratio is 5x . Therefore the largest Angle is ,

=> Largest Angle = 5x

=> Largest Angle = 5 × 20°

=> Largest Angle = 100°

<h3>• Hence the largest Angle is 100° </h3>

7 0

3 years ago

Hard question but would help me out

Alekssandra [29.7K]

Answer:

Critical value f(1)=2.

Minimum at (1,2), function is decreasing for $0$ and increasing for $x>1.$

$\left(3,\dfrac{4}{\sqrt{3}}\right)$ is point of inflection.

When 0<x<3, function is concave upwards and when x>3, , function is concave downwards.

Step-by-step explanation:

1. Find the domain of the function f(x):

$\left\{\begin{array}{l}x\ge 0\\x\neq 0\end{array}\right.\Rightarrow x>0.$

2. Find the derivative f'(x):

$f'(x)=\dfrac{(x+1)'\cdot \sqrt{x}-(x+1)\cdot (\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}=\dfrac{2x-x-1}{2x\sqrt{x}}=\dfrac{x-1}{2x^{\frac{3}{2}}}.$

This derivative is equal to 0 at x=1 and is not defined at x=0. Since x=0 is not a point from the domain, the crititcal point is only x=1. The critical value is

$f(1)=\dfrac{1+1}{\sqrt{1}}=2.$

2. For $0$ the derivative f'(x)<0, then the function is decreasing. For $x>1,$ the derivative f'(x)>0, then the function is increasing. This means that point x=1 is point of minimum.

3. Find f''(x):

$f''(x)=\dfrac{(x-1)'\cdot 2x^{\frac{3}{2}}-(x-1)\cdot (2x^{\frac{3}{2}})'}{(2x^{\frac{3}{2}})^2}=$

$=\dfrac{2x^{\frac{3}{2}}-2(x-1)\cdot \frac{3}{2}x^{\frac{1}{2}}}{4x^3}=\dfrac{2x^{\frac{3}{2}}-2\cdot\frac{3}{2}x^{\frac{3}{2}}+ 2\cdot\frac{3}{2}x^{\frac{1}{2}}}{4x^3}=$

$=\dfrac{-x+3}{4x^{\frac{5}{2}}}.$

When f''(x)=0, x=3 and $f(3)=\dfrac{3+1}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}.$

When 0<x<3, f''(x)>0 - function is concave upwards and when x>3, f''(x)>0 - function is concave downwards.

Point $\left(3,\dfrac{4}{\sqrt{3}}\right)$ is point of inflection.

4 0

4 years ago

(5a-2)(5a+2)-3(1-a)-3a=

attashe74 [19]

So first we need to know the order of operations
P - Parenthesis
E - Exponents
M - Multiplication
D - Division
A - Addition
S - Subtraction

Then the special cases:
(I have attached a picture below)

Lets get into action!
(5a - 2)(5a + 2) - 3( 1 - a ) - 3a
25a² - 4 - 3 + 3a - 3a
25a² - 7
25a² - 7 is the answer