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tankabanditka [31]
3 years ago
11

Which of these options is not a quadratic equation in x?

Mathematics
2 answers:
il63 [147K]3 years ago
6 0

Answer:

3x³ - 2x² + 1 = 0

Step-by-step explanation:

By definition, a quadratic equation cannot have an exponent higher than 2.

ludmilkaskok [199]3 years ago
5 0

Answer:

3x³ - 2x² + 1 = 0

Step-by-step explanation:

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You scored 31 out of 40 points on a test. What percent did you get?
elena-14-01-66 [18.8K]

Answer:


Step-by-step explanation:

Divide 31 by 40. Since you get .775, you make it a percent and it's 77.5% correct.

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3 years ago
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Step-by-step explanation:

Equation:

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2 years ago
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Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0
Leokris [45]

Substitute v(x)=-2x+y(x), so that \dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}. Then the ODE is equivalent to

\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7

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\dfrac{\mathrm dv}{v^2-9}=\mathrm dx

Split the left side into partial fractions,

\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)

so that integrating both sides is trivial and we get

\dfrac{\ln|v-3|-\ln|v+3|}6=x+C

\ln\left|\dfrac{v-3}{v+3}\right|=6x+C

\dfrac{v-3}{v+3}=Ce^{6x}

\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}

\dfrac6{v+3}=1-Ce^{6x}

v=\dfrac6{1-Ce^{6x}}-3

-2x+y=\dfrac6{1-Ce^{6x}}-3

y=2x+\dfrac6{1-Ce^{6x}}-3

Given the initial condition y(0)=0, we find

0=\dfrac6{1-C}-3\implies C=-1

so that the ODE has the particular solution,

\boxed{y=2x+\dfrac6{1+e^{6x}}-3}

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3 years ago
Find the slope and y-intercepts <br> f(x)= -1/2x - 2
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Answer:

slope: -1/2

y-intercept: -2

Step-by-step explanation:

f(x) = mx+b

m is slope

b is the y-intercept

so

if f(x) = -1/2x - 2

then

the slope or m is -1/2

and the y-intercept or b is -2

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3 years ago
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Veronika [31]

Answer:

Step-by-step explanation:

the area is L x W so its 42cm

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