All categories

2 years ago

Area of the shaded segment=

Mathematics

1 answer:

garik1379 [7]2 years ago

5 0

The area of the entire sector is $(\pi)(6^{2}) \left(\frac{60}{360} \right)=6\pi$

The area of the triangle OAB is $\frac{\sqrt{3}}{4}(6^{2})=9\sqrt{3}$ .

So, the area of the segment is $\boxed{6\pi-9\sqrt{3}}$

You might be interested in

What is theory of demand?

Yuri [45]

Answer:

Demand theory is an economic principle relating to the relationship between consumer demand for goods and services and their prices in the market. ... As more of a good or service is available, demand drops and so does the equilibrium price.

4 0

3 years ago

Read 2 more answers

Determine if the function is one-to-one.

olya-2409 [2.1K]

Answer:

Step-by-step explanation:

From what I saw on google for what "one to one" means, It's basically meaning that the bottom of that quadratic function is at point (0 , 0) and hits both (-1 , 1) and (1 , 1) on the graph.

4 0

3 years ago

Six different second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic re

Sauron [17]

Answer:

What? Hey no no no yes what no no yes what yes yes

7 0

3 years ago

Which property does the following equation represent? 62? + 43 = 43+ 6x2

timurjin [86]

Answer:

commutative

Step-by-step explanation:

you are typing the same numbers but in different order

8 0

3 years ago

Please answer correctly !!!!!!!!! Will Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

6 0

3 years ago