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Kipish [7]
3 years ago
9

Please help will give brainlist!​

Mathematics
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

b=\frac{1}{6}

Step-by-step explanation:

In f(x)=a\cos(b(x-c))+d, b represents a constant related to the period of the function. Here's how it's related:

b=\frac{2\pi}{T}, where T is the period of the function.

We're given T=12\pi, so solving for b:

b=\frac{2\pi}{12\pi}=\boxed{\frac{1}{6}}

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I neeed help asap please
alexandr1967 [171]

Answer:

last option.

linear pair

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6 0
3 years ago
Solve the following equation for x<br> 2x+6=3x−12x+6=3x−1
Rudik [331]

Answer: x=11/22

Step-by-step explanation: 2x+6=3x−12x+6=3x-1                                           2x+6=3x+−12x+6=3x+-1                                                                                       2x+6=(3x+−12x)+(6+-1)=3x                                                                     2x+6+9x=−9x+5+9x=3x+9x                                                                                   11x+6=5=12x                                                                                                 11x+12x=6+5                                                                                             22x/22=11/22                                                                                                       x=11/22                                                                                                                                                                                                                                                                                            

3 0
3 years ago
Help me with these 2 questions please
pishuonlain [190]

Answer:

I believe for the second one it's B, then for the third one it's D

Step-by-step explanation:

Oh also be careful w ur lunch # showing

4 0
2 years ago
How do you do question b?
notsponge [240]

Part (b)

We use the result of part (a) and plug in (x,y) = (0,0). This is directly from the initial condition y(0) = 0.

\arcsin(4y) = x^2 + C\\\\\arcsin(4*0) = (0)^2 + C\\\\\arcsin(0) = C\\\\0 = C\\\\C = 0\\\\

-----------------

This means,

\arcsin(4y) = x^2 + C\\\\\arcsin(4y) = x^2 + 0\\\\\arcsin(4y) = x^2\\\\4y = \sin(x^2)\\\\y = \frac{1}{4}\sin(x^2)\\\\

is the solution with the initial condition y(0) = 0.

6 0
3 years ago
ANSWER QUICKLY PLEASEEEE
Nitella [24]

Answer:

it would be c

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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