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3 years ago

Which equation describes a relationship such that x = 0 must be excluded from the domain?

Mathematics

2 answers:

Ilia_Sergeevich [38]3 years ago

5 0

y = 2 / x

the answer cant have 0 : )

please rate and thank : )

dem82 [27]3 years ago

3 0

Y = 2 / x
Cannot have x = 0 in its domain because that will cause the denominator to be 0, making the value of y undefined.

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missed my lecture and need help with this derivative, I tried looking at online resources but I just got confused so if someone

Zigmanuir [339]

Derivative with respect to x: $12x^2y^2-\frac{z^3}{x}+16x^{15}$

Derivative with respect to y: $8x^3y-4e^zy^3+4$

Derivative with respect to z: $-e^z y^4 +\frac{3z^2}{x^2}$

Step-by-step explanation:

We can calculate three derivatives for this function: the derivative with respect to x, the derivative with respect to y, and the derivative with respect to z.

We start with the derivative with respect to x:

$\frac{df}{dx}=4(3x^{3-1})y^2 -0 +z^3(-x^{-1})+0-16x^{16-1}=\\=12x^2y^2-\frac{z^3}{x}+16x^{15}$

Then we calculate the derivative with respect to y:

$\frac{df}{dy}=4x^3(2y^{2-1})-e^z(4y^{4-1})+0+4y^{1-1}=8x^3y-4e^zy^3+4$

Finally, we calculate the derivative with respect to z:

$\frac{df}{dz}=0-e^z y^4 +\frac{3z^{3-1}}{x^2}+0-0=-e^z y^4 +\frac{3z^2}{x^2}$

Learn more about differentiation:

brainly.com/question/4279146

#LearnwithBrainly

5 0

3 years ago

Mary Beth wants to save $175 for a trip to an amusement park. She sets aside $14 of her allowance at the end of each week. How m

scoray [572]

Answer:

Approximately 13 weeks

Step-by-step explanation:

GIven data

Amount to be save= $175

Weekly allowance = $14

Hence the number of weeks it will take to save $175 is given as

=175/14

=12.5

Therefore the number of weeks is approximately 13 weeks

3 0

2 years ago

Prove identity trigonometric equation

irina [24]

Explanation:

The given equation is False, so cannot be proven to be true.

Perhaps you want to prove ...

$2\tan{x}=\dfrac{\cos{x}}{\csc{(x)}-1}+\dfrac{\cos{x}}{\csc{(x)}+1}$

This is one way to show it:

$2\tan{x}=\cos{(x)}\dfrac{(\csc{(x)}+1)+(\csc{(x)}-1)}{(\csc{(x)}-1)(\csc{(x)}+1)}\\\\=\cos{(x)}\dfrac{2\csc{(x)}}{\csc{(x)}^2-1}=2\cos{(x)}\dfrac{\csc{x}}{\cot{(x)}^2}=2\dfrac{\cos{(x)}\sin{(x)}^2}{\cos{(x)}^2\sin{(x)}}\\\\=2\dfrac{\sin{x}}{\cos{x}}\\\\2\tan{x}=2\tan{x}\qquad\text{QED}$