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2 years ago

14

What is the exponential regression equation to best fit the data where the y values were rounded to the nearest integer.

1 answer:

ladessa [460]2 years ago

6 0

Answer:

the answer is Y=###########^x

Step-by-step explanation:

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Simplify 6 - 3(4 - 3x)

IRISSAK [1]

<span>6 - 3(4 - 3x)
= 6 - 12 + 9x
= 9x - 6

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2 years ago

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Ming writes a number pattern on a slip of paper and hands it to his friend Jack. 24 21 23 20 22 19 21 18 Jack writes the next nu

tiny-mole [99]

The rule is -3, then +2

24-3 = 21
21 + 2 = 23
23-3= 20
20+2= 22
22-3= 19
19+2 = 21
21-3= 18
18+2 = 20

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2 years ago

PLEASE HELP!!!!!! ILL MAKE BRAINLIEST

lesya692 [45]

Answer:

x = 17

Step-by-step explanation:

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2 years ago

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Simplify: x0 y-3 / x2 y-1

svetlana [45]

$\frac{x {}^{0}y {}^{ - 3} }{x {}^{2}y {}^{ - 1} } \\ = \frac{1y {}^{ - 3} }{ {x}^{2}y {}^{ - 1} } \\ = \frac{1 \times 1}{ {x}^{2} y {}^{2} } \\ \\ = \frac{1}{ {x}^{2}y {}^{2} }$

Step by Step Explanation:

Evaluate the power: Any non-zero expression raised to the power of 0 equals 1
Simplify: Simplify the expression
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<h2>☆彡Hanna</h2>

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8 0

2 years ago

Please I need help with differential equation. Thank you

Inga [223]

1. I suppose the ODE is supposed to be

$\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)$

Solving for $\dfrac{\mathrm dy}{\mathrm dt}$ gives

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}$

which is undefined when $t=\pm1$ . The interval of validity depends on what your initial value is. In this case, it's $t=-\dfrac12$ , so the largest interval on which a solution can exist is $-1\le t\le1$ .

2. Separating the variables gives

$\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}$

Integrate both sides. On the left, we have

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}$

where we substituted $z=y^{1/2}$ - or $z^2=y$ - and $2z\,\mathrm dz=\mathrm dy$ - or $\mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}$ .

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)$

On the right, we have

$\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)$

$\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C$

So

$2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C$

$\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C$

$y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}$

$y^{1/2}=C(1-t^2)^{1/4}-1$

I'll leave the solution in this form for now to make solving for $C$ easier. Given that $y\left(-\dfrac12\right)=1$ , we get

$1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1$

$2=C\left(\dfrac54\right)^{1/4}$

$C=2\left(\dfrac45\right)^{1/4}$

and so our solution is

$\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}$

3 0

3 years ago