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3 years ago

If the function m(x) goes

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

7 0

Answer:

(-5,15)

Step-by-step explanation:

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How do I solve <br><img src="https://tex.z-dn.net/?f=%20%20%5Csin%28%20a%20%29%20%20%3D%20.8325" id="TexFormula1" title=" \sin(

Serga [27]

Answer:

$a=56.4\degree$

Step-by-step explanation:

Let $a$ be an acute angle.

To solve: $\sin(a)=0.8325$ , we take the sine inverse of both sides to obtain:

$\sin^{-1}(\sin(a))=\sin^{-1}(0.8325)$

Recall composition property of a function and its inverse function.

$f^{-1}(f(x))=x$

We apply this property to the left hand side to obtain;

$a=\sin^{-1}(0.8325)$

We now use our scientific calculator to obtain;

$a=56.3564115\degree$

We round to the nearest tenth to obtain;

$a=56.4\degree$

4 0

3 years ago

Can someone pls help me

Eduardwww [97]

Answer:

Step-by-step explanation:

I hope that is useful for you

8 0

2 years ago

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

Lina20 [59]

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

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

7 0

3 years ago

When recording a transaction in a journal, the account listed first is always the:

GalinKa [24]

Answer:

da te

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please help explanation if possible

egoroff_w [7]

Answer: x = 2 and y = -4

x + 2y = -6

x = -6-2y

Putting this in value of x in

6x + 2y = 4

6(-6-2y) + 2y = 4

-36-12y+2y = 4

-10y = 4+36

y = 40/(-10)

y = -4

Now putting this value of y in

x + 2y = -6

x + 2(-4) = -6

x -8 = -6

x = -6+8

x = 2

Therefore x = 2 and y = -4

If we put these values we can check this

x + 2y = -6

2 + 2(-4) = -6

2 -8 = -6

-6 = -6

please click thanks and mark brainliest if you like :)

7 0

2 years ago

Read 2 more answers