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Sholpan [36]
2 years ago
8

Helpppppppppppp meeeeeee

Mathematics
1 answer:
MArishka [77]2 years ago
4 0

Answer:

7

Step-by-step explanation:

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Help me pls i need help asap​
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Step-by-step explanation:

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Todea what did one pencil say to another mathematics algebra order of operations answer
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Part 1

6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)}  \\  \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\  \\ =3



Part 2

7-8\div4(3^2-1)=7-8\div4(9-1) \\  \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\  \\ =6  \frac{1}{2} = \frac{13}{2}



Part 3

2-2(2-5)^2+3^2=2-2(-3)^2+9 \\  \\ =2-2(9)+9=2-18+9=-7



Part 4

(5^2-9\cdot3)^2-11=(25-27)^2-11 \\  \\ =(-2)^2-11=4-11=-7



Part 5

[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\  \\ =(1-5)\cdot4=-4\cdot4=-16



Part 6

12-11\cdot2+16\div8=12-22+2 \\  \\ =-8



Part 7

-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\  \\ =-2-(-1)=-2+1=-1



Part 8

8+2\cdot3-14\div7=8+6-2 \\  \\ =12



Part 9

(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\  \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\  \\ =36+4=40



Part 10

7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\  \\ =7+(-7)\cdot5=7+(-35)=7-35 \\  \\ =-28



Part 11

3-2[2^3+(-1)^3]=3-2[8+(-1)] \\  \\ =3-2(8-1)=3-2(7)=3-14 \\  \\ -11



Part 12

18+6\div3(2^2+5)=18+6\div3(4+5) \\  \\ =18+6\div3(9)=18+6\div27=18+ \frac{2}{9} \\  \\  =18 \frac{2}{9}
8 0
3 years ago
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
Solve for k. 3/4 k+3/8 k=12
Alexandra [31]
I hope this helps you

8 0
3 years ago
Read 2 more answers
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