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antiseptic1488 [7]

3 years ago

5

Each student earns 3/4 point for every bonus question. If a student gets 6 bonus points, how many bonus questions did a student

answer correctly?

2 answers:

galina1969 [7]3 years ago

8 0

Answer:

8

Step-by-step explanation:

klio [65]3 years ago

6 0

Answer:

8

Step-by-step explanation:

6 divided by.75 if u think I'm wrong check and correct me

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The basic price for an annual dental checkup and cleaning is $50. If the dentist finds any cavities in there will be an addition

Slav-nsk [51]

Answer:

100n+50

Step-by-step explanation:

the dentist charges $100 dollars per cavity found. Per indicates multiplication. Since there is an additional fee, we add $50 to the $100n, which will make the expression

<em>$</em><em>100n</em><em>+</em><em>5</em><em>0</em>

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3 years ago

I’ll give points + brainslist, if you don’t know don’t bother answering (:

Damm [24]

C. 35 because c. Is the right answer just trust me ok

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4 years ago

Which graph shows a linear function?

12345 [234]

Answer:

The first graph, upper left corner

Step-by-step explanation:

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4 years ago

A. Find the slope of the line that passes through the points (2,-5) and (-2,3).

Crazy boy [7]

Answer:

a) Slope: -2

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Step-by-step explanation:

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5 0

3 years ago

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$

so the new equation is

$y''(t)+ 8y'(t)-20y(t)=0$

the auxiliary equation is

$x^2+8x-20=0\\ x^2-2x+10x-20=0\\x(x-2)+10(x-2)=0\\(x+10)(x-2)=0\\ x=-10\text{ or }x=2$

so the solutions of the new equation are

$y(t)=ae^{2t}+be^{-10t}$

with a and b real

as

$x(t)=e^t\\ t(x)=ln(x)$

$y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}$

hope this helps

do not hesitate if you have any questions

8 0

4 years ago