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Ilia_Sergeevich [38]
3 years ago
13

25 points/ brainlyest help plzzzz check the photo

Mathematics
1 answer:
Oxana [17]3 years ago
6 0

I think it is a compressed spring.


hope it helps

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Write an equation for the line shown on the right.<br> The equation of the line is y
Alenkasestr [34]

Answer:

y= -14x +5

I hope this helps :)

7 0
3 years ago
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Val returned a bicycle while she was on vacation. She paid a flat rental fee of $55.00 plus $8.50 each day. The total cost was $
andrew-mc [135]

Answer:

8 days

Step-by-step explanation:

The total cost includes $55 flat rental fee which is one time cost only as not given everyday so we subtract it from $123 and divide $(123-55) with per day cost that is $8.50 which gives 8 as answer so Val spent 8 days on vacation.

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3 years ago
3[-x+(2x+1)]=x-1 ? What is X?
Nonamiya [84]
<span>3[-x+(2x+1)]=x-1
</span><span>3[x+1]=x-1
3x + 3 = x -1
3x - x = -1 -3
2x = -4
x = -4/2
x = -2</span>
7 0
3 years ago
You are given the equation 27bx7•cx8•7xd=455x27. If b and c are both integers greater than 1 and b less than cb
jolli1 [7]

Answer:

The question is not so clear and complete

Step-by-step explanation:

But for questions like this, since the equation has been given, what is expected is for us to make comparison, compare the RHS with the LHS or by method of comparing coefficients.

We follow the stated conditions since we are told that b and c are both integers which are greater than 1 and b is less than the product of cb. from these conditions, we can compare and get the values of b , c and d.

Another approach is to assume values, make assumptions with the stated conditions, however, our assumptions must be valid and correct if we substitute the assumed values of b, c and d in the equation, it must arrive at the same answer for the RHS. i.e LHS = RHS

5 0
3 years ago
Find the derivatives of the following implicit function
ddd [48]

Answer:

\frac{d}{dx}\left(y\right)=\frac{2-6x+6y}{-6x+2y+1}

Step-by-step explanation:

3x^2-6xy+y^2=2x-y\\\mathrm{Treat\:}y\mathrm{\:as\:}y\left(x\right)\\\mathrm{Differentiate\:both\:sides\:of\:the\:equation\:with\:respect\:to\:}x\\\frac{d}{dx}\left(3x^2-6xy+y^2\right)=\frac{d}{dx}\left(2x-y\right)\\\frac{d}{dx}\left(3x^2-6xy+y^2\right)=6x-6\left(y+x\frac{d}{dx}\left(y\right)\right)+2y\frac{d}{dx}\left(y\right)\\\frac{d}{dx}\left(2x-y\right)=2-\frac{d}{dx}\left(y\right)\\6x-6\left(y+x\frac{d}{dx}\left(y\right)\right)+2y\frac{d}{dx}\left(y\right)=2-\frac{d}{dx}\left(y\right)

\mathrm{For\:convenience,\:write\:}\frac{d}{dx}\left(y\right)\mathrm{\:as\:}y^{'\:}\\6x-6\left(y+xy^{'\:}\right)+2yy^{'\:}=2-y^{'\:}\\\mathrm{Isolate}\:y^{'\:}:\quad y^{'\:}=\frac{2-6x+6y}{-6x+2y+1}\\y^{'\:}=\frac{2-6x+6y}{-6x+2y+1}\\\mathrm{Write}\:y^{'\:}\:\mathrm{as}\:\frac{d}{dx}\left(y\right)\\\frac{d}{dx}\left(y\right)=\frac{2-6x+6y}{-6x+2y+1}

3 0
3 years ago
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