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baherus [9]
3 years ago
9

Determine whether the graph table or equation represents a linear nonlinear function.

Mathematics
1 answer:
Fofino [41]3 years ago
3 0

Answer:

nonlinear

Step-by-step explanation:

x(x + 3) =  {x}^{2}  + 3  \\ \: it \: will \: be \: a \: parabola \: because \: of \: the \:  {x}^{2}

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Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s
RSB [31]
Well, is the function continuous?  yes, is a cubic one, you can graph it if you wish, is continuous all the way, and of course at [ 0, 4] too.

is it differentiable?  you can always look at the graph between 0,4 and is a smooth transition line, thus yes, it is differentiable, but let's check anyway,

\bf \cfrac{dy}{dx}=3x^2-2x-12    its derivative has no asymptotes and therefore no "cusps", so yes, is differentiable all around.

is f(0) = f(4), let's check

f(0) = 0+0+0+3,            f(0) = 3
f(4) = 64 - 16 - 48 + 3,   f(4) = 3

yeap

there must then be a "c" value(s) with a horizontal tangent slope, let's check, is really just the critical points.

\bf \cfrac{dy}{dx}=3x^2-2x-12\implies 0=3x^2-2x-12
\\\\\\
\textit{using the quadratic formula}
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x=\cfrac{-(-2)\pm\sqrt{(-2)^2-4(3)(-12)}}{2(3)}\implies x=\cfrac{2\pm\sqrt{4+144}}{6}
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x=\cfrac{2\pm 2\sqrt{37}}{6}\implies x=\cfrac{2\pm\sqrt{37}}{3}\impliedby \textit{c's}
4 0
3 years ago
What is the third term of the sequence?<br> A<br> 25<br> B<br> 50<br> C. 35<br> D<br> 20
hjlf

Answer:

C

Step-by-step explanation:

35 is the third term of the sequence

Have a great day

8 0
3 years ago
If the volume of the crystal ball is about 113.14 cm3, the radius of the ball is approximately
stellarik [79]

Answer:

r ≅ 3.00

Step-by-step explanation:

First we need to get the formula for finding the volume of sphere. The formula  is:

V=\frac{4}{3}\pi r^{3}

Now to find for the radius, we need to input the value that we know into the formula.

113.14=\frac{4}{3}\pi r^{3}

Now we multiply both sides by 3/4.

\frac{3}{4}x113.14=\frac{4}{3}\pi r^{3}

84.86=\pi r^{3}

Now divide both sides by π.

\frac{84.86}{\pi}=\frac{\pi r^{3}}{\pi}

27.01=r^{3}

To find r, we finally get the cube root of both sides.

\sqrt[3]{r^{3} }= \sqrt[3]{27.01}

r ≅ 3.00

4 0
3 years ago
Read 2 more answers
Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^
serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'

=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'

=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'

=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}

=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}

Note that writing \log(\ln x)^3=3\log(\ln x) is one way to avoid using the power rule.

(2)

\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'

=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'

=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}

=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}

(3)

\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'

=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')

=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})

=3x^2e^{3x}(1+x)\cos(x^3e^{3x})

(4)

\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'

=3\sin^2(xe^x)\cos(xe^x)(xe^x)'

=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')

=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)

=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)

(5) Use implicit differentiation here.

(\ln(xy))'=(e^{2y})'

\dfrac{(xy)'}{xy}=2e^{2y}y'

\dfrac{x'y+xy'}{xy}=2e^{2y}y'

y+xy'=2xye^{2y}y'

y=(2xye^{2y}-x)y'

y'=\dfrac y{2xye^{2y}-x}

8 0
3 years ago
45678 ÷ 30 (Please complete all the steps in the notebook)
Vikentia [17]

Answer:

`1522.6

Step-by-step explanation:

See image below:)

3 0
3 years ago
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