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2 years ago

8

How do you simplify √216

1 answer:

svp [43]2 years ago

4 0

$\sqrt{216} =\sqrt{36 \times 6} =\boxed{\bf{6\sqrt{6}}}$

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Pls answer the question attached

barxatty [35]

$\huge\mathfrak{\underline{answer:}}$

$\large\bf{\angle ACD = 105°}$

__________________________________________

$\large\bf{\underline{Here:}}$

BCD is an isosceles right triangle , right angled at D
ABC is an equilateral triangle

$\large\bf{\underline{To\:find:}}$

∠ ACD

$\large\bf{In\: triangle\:ABC:}$

❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same

$\large\bf{\underline{So}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = \frac{180}{3}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = 60°}$

$\large\bf{In\: triangle\:BDC}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle D = 90°}$

$\bf{⟼\angle DBC =\angle DCB }$ (Isosceles triangle)

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB + 90° = 180°}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB = 180 -90}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = \frac{90}{2}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = 45°}$

$\large\bf{\underline{Therefore,}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = \angle ACB + \angle DCB}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = 60+45}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟹\angle ACD = 105°}$

6 0

3 years ago

A certain radioactive material decays in such a way that the mass in kilograms remaining after t years is given by the function

Angelina_Jolie [31]

The mass of radioactive material remaining after 50 years would be 48.79 kilograms

<h3>How to determine the amount</h3>

It is important to note that half - life is the time it takes for the amount of a substance to reduce by half its original size.

Given the radioactive decay formula as

m(t)=120e−0.018t

Where

t= 50 years

m(t) is the remaining amount

Substitute the value of t

$m(t) = 120e^-^0^.^0^1^8^(^5^0^)$

$m(t) = 120e^-^0^.^9$

Find the exponential value

m(t) = 48.788399

m(t) = 48.79 kilograms to 2 decimal places

Thus, the mass of radioactive material remaining after 50 years would be 48.79 kilograms

Learn more about half-life here:

brainly.com/question/26148784

#SPJ1

7 0

1 year ago

Free brainliest for first to answerrr . check my questions on my profile for more like this !

Elena L [17]

Am I supposed to answer anything lol

6 0

2 years ago

Read 2 more answers

anyone know the answer i worked it out but didn't know where i messed up so i am just gonna ask for someone to help me with it.

hram777 [196]

I also worked it out but I got -5/3

2(6p-5) ≥ 3(p-8) -1

12p-10 ≥ 3p-24-1

12p-10 ≥ 3p-25

12p-3p ≥ -25+10

9p ≥ -15

-15/9

P ≥ -5/3

5 0

2 years ago

What is the area of a sector with a central angle of 108 degrees and a diameter of 21.2 cm?

Rasek [7]

Answer:

$105.84\ cm^2$

Step-by-step explanation:

step 1

Find the area of the circle

The area of the circle is

$A=\pi r^{2}$

we have

$r=21.2/2=10.6\ cm$ ----> the radius is half the diameter

substitute

$A=\pi (10.6)^{2}$

$A=112.36\pi\ cm^2$

step 2

Find the area of a sector

Remember that

The area of the circle subtends a central angle of 360 degrees

so

using proportion

Find out the area of a sector with a central angle of 108 degrees

$\frac{112.36\pi}{360^o}=\frac{x}{108^o}\\\\x=112.36\pi(108)/360\\\\x=33.708\pi\ cm^2$

assume

$\pi=3.14$

substitute

$33.708(3.14)=105.84\ cm^2$

7 0

3 years ago