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2 years ago

7 1/7 divided by 10/11 simplify

Mathematics

1 answer:

Klio2033 [76]2 years ago

6 0

Exact Form:
620
77
620
77
Decimal Form:
8.05194805
…
8.05194805
Mixed Number Form:
8
4
77

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The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
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\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
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\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
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\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
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\\y' = \frac{3}{( -32 )}$

7 0

2 years ago

The sum of 5 consecutive integers is 265. What is the fifth number in this sequence ?

Yanka [14]

Hello,
Let's assume n,n+1,n+2,n+3,n+4 the 5 numbers

n+(n+1)+...+(n+4)=5n+10=265
5n=265-10
5n=255
n=51
The 5th number is 51+4=55

6 0

2 years ago

6(x+7)x=9<br><br> A.12<br> B.42<br> C.96<br> D.16<br><br> Please work out the problem.

marin [14]

X = - $\frac{7}{2}$ + $\frac{1}{2} \sqrt{55}$ or x = - $\frac{7}{2}$ + - $\frac{1}{2} \sqrt{55}$

4 0

2 years ago

Help PLEASE (look at picture)

elena-14-01-66 [18.8K]

Answer:

Step-by-step explanation:

So for QR is 1-(-3)=4

QT:|-2-2|=4

TS: |-2-(-2)|=4

RS: 2-(-2)=4

and they plus together is 4x

4x4 = 16

I hope this helps, have a blessed day. :D

6 0

1 year ago

Which trigonometric functions are negative in the fourth (IV) quadrant?

anzhelika [568]

<h3>Answers are: sine, tangent, cosecant, cotangent</h3>

Explanation:

On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.

Since sine is negative, so is cosecant as this is the reciprocal of sine

csc = 1/sin

In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.

Because tangent is negative, so is cotangent.

The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.

7 0

2 years ago