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3 years ago

5

What is the quadratic regression equation that fits these data?

1 answer:

ratelena [41]3 years ago

4 0

Answer:

y = 2.09x2 + 0.33x + 3.06

Step-by-step explanation:

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Car decreased in price by 3%. Original price is $45600. New Price= A x original price

Karo-lina-s [1.5K]

A decrease of 3% means the car is now valued at 97% of the original price. ( 100% -3% = 97%)

multiply original price by 97%

45,600 x 0.97 = 44,232

Answer: $44,232

7 0

2 years ago

There is a fraction equivalent to 6/7. The sum of the numerator and the denominator is 78. What is the fraction

Zanzabum

The fraction is 36/42.

36 + 42 = 78
36/42 = 6/7

4 0

4 years ago

Consider the following higher-order differential equation. d3u dt3 + d2u dt2 − 2u = 0 Find all the roots of the auxiliary equati

g100num [7]

Answer:

Therefore the auxiliary solution is

$y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]$

Step-by-step explanation:

To find auxiliary equation we have to put $u=e^{mt}$ in the given differential equation.

The degree of the differential equation is 3.

Therefore the number of root of the differential equation is 3.

Let $\lambda_1$ , $\lambda_2$ and $\lambda_3$ be three roots of the auxiliary equation.

If three roots are real and equal.

Then $y= e^{\lambda_1t} (C_1+C_2t+C_3t^2)$

If three roots are real and distinct.

Then $y=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}+C_3e^{\lambda_3x}$

If two roots imaginary and one root real , $\lambda_1= a+ib \ and \ \lambda_2= a-ib$

Then $y=C_1e^{\lambda_3t}+e^{at}(C_2sin \ bt+C_3cos \ bt)$

Now, $u=e^{mt}$ , $u'=me^{mt},u"=m^2e^{mt} \ and \ u'"= m^3e^{mt}$

Given differential equation is

$\frac{d^3u}{dt^3} +\frac{d^2u}{dt^2}-2u=0$

The auxiliary equation is

$(m^3+m^2-2)e^{mt}=0$

$\Rightarrow (m^3+m^2-2)=0$

$\Rightarrow m^3-m^2+2m^2-2m+2m-2=0$

$\Rightarrow m^2(m-1)+2m(m-1)+2(m-1)=0$

$\Rightarrow (m-1)(m^2+2m+2)=0$

$\Rightarrow m= 1,\frac{-2\pm\sqrt{2^2-4.2.1} }{2.1}$

$\Rightarrow m= 1,\frac{-2\pm\sqrt{-4} }{2.1}$

$\Rightarrow m= 1,-1\pm i$

Here $\lambda_1= -1+i , \lambda_2=- 1-i \ and \ \lambda_3=1$

Therefore ,

$y=C_1e^{1.t}+e^{-1.t}[C_2sin \ (1.t)+C_3 cos \ (1.t)]$

$\Rightarrow y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]$

5 0

4 years ago

What is y=(x-1)²+2

Andrei [34K]

The answer is no x-intercept/zero.

4 0

3 years ago

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A mountain climber ascended 3/4 of a kilometer in 1/12 of an hour. What was the mountain climber’s ascent rate in kilometers per

Yanka [14]

<span>A mountain climber ascended 3/4 of a kilometer in 1/12 of an hour. What was the mountain climber’s ascent rate in kilometers per hour?

Correct Answer: C. 9 Kilometers per hour </span>

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3 years ago

Read 2 more answers