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Tanya [424]
2 years ago
8

Hello guysssssss asap

Mathematics
1 answer:
Vikentia [17]2 years ago
8 0

Answer:

-9, -8, -4 , 6 then 9..

Step-by-step explanation:

Your welcome

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3 years ago
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There are six wires which need to be attached to a circuit board. A robotic device will attach the wires. The wires can be attac
lesya692 [45]

Answer:

S=720

Step-by-step explanation:

From the question we are told that

Six wire are attached to a circuit board

Generally using multiplication rule

Let number of possible sequence be given as S

Mathematically  

S=6! =6*5*4*3*2*1=720

Therefore

Number of possible sequence S=720

5 0
2 years ago
The excluded values of a rational expression are 2 and 5. Which of the following could be this expression?
LuckyWell [14K]

Answer:

You have not added the options, therefore, I cannot provide an exact answer. However, I can help you with the concept.

In any fraction, the denominator cannot be equal to zero because this would make the fraction undefined.

Excluded values are the values that would make the denominator equal to zero.

We are given that the excluded values are 2 and 5, this means that the factors of the denominator are (x-2) and (x-5)

This means that the denominator is x² - 7x + 10

Pick the fraction which has this denominator.

Hope this helps :)

3 0
3 years ago
Please help I’ll give brainliest
AlekseyPX
It’s the 2nd on I’m pretty sure because it says decimal
8 0
2 years ago
Read 2 more answers
PLEASE HELP!!
erik [133]
2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha

1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}

3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x
8 0
2 years ago
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