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3 years ago

8

WILL GIVE BRAINLEIST Which graph shows three points that represent equivalent ratios?

2 answers:

Margaret [11]3 years ago

8 0

Answer:

The answer is c.

Step-by-step explanation: i took the test

im 100% sure

Inessa [10]3 years ago

4 0

Answer: The first one and the last one

Step-by-step explanation:

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If your little sister watches you and tries to copy everything you do, she is exhibiting which type of learning?

Elis [28]

<span>social learning

</span>If your little sister watches you and tries to copy everything you do, she is exhibiting which type of learning?
NOT:
classical conditioning
insightful learning
<span>habituation</span>

8 0

3 years ago

What is the equation of a circle with center (−8, 3) and radius 8?

Genrish500 [490]

Answer:

x^2 + y^2 + 16x + 6y + 9 = 0

Step-by-step explanation:

Using the formula for equation of a circle

(x - a)^2 + (y + b)^2 = r^2

(a, b) - the center

r - radius of the circle

Inserting the values given in the question

(-8,3) and r = 8

a - -8

b - 3

r - 8

[ x -(-8)]^2 + (y+3)^2 = 8^2

(x + 8)^2 + (y + 3)^2 = 8^2

Solving the brackets

( x + 8)(x + 8) + (y +3)(y+3) = 64

x^2 + 16x + 64 + y^2 + 6y + 9 = 64

Rearranging algebrally,.

x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0

Bringing in 64, thereby changing the + sign to -

Therefore, the equation of the circle =

x^2 + y^2 + 16x + 6y + 9 = 0

7 0

3 years ago

What is the slope of this line

prisoha [69]

Answer:

don't bjjjjjjjjj*jjjjjjjjjjj

8 0

3 years ago

1. A woman deposits $10000 in a bank. The interest rate per annum is 8% How much will she have in the account in 15 years to the

jeka57 [31]

Answer:

The answer is $31,722.

Step-by-step explanation:

10,000 x 1.08 = 10,800

Do this 14 more times, your answer times 1.08.

You end up with 31,721.6911418.

Rounded to the nearest dollar, that would be $31,722.

4 0

2 years ago

Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4

OLEGan [10]

Answer:

$y(0.2)=3$ , $y(0.4)=3.005974448$ , $y(0.6)=3.017852169$ , $y(0.8)=3.035458382$ , and $y(1.0)=3.058523645$

The general solution is $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

The error in the approximations to y(0.2), y(0.6), and y(1):

$|y(0.2)-y_{1}|=0.002982771$

$|y(0.6)-y_{3}|=0.008677796$

$|y(1)-y_{5}|=0.013499859$

Step-by-step explanation:

<em>Point a:</em>

The Euler's method states that:

$y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)$ where $t_{n+1}=t_n + h$

We have that $h=0.2$ , $t_{0}=0$ , $y_{0} =3$ , $f(t,y)=3te^{-y}$

We need to find $y(0.2)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{1}=t_{0}+h=0+0.2=0.2$

$y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=$

$=3 + 0.2 \cdot \left(0 \right)= 3$

$y(0.2)=3$

We need to find $y(0.4)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{2}=t_{1}+h=0.2+0.2=0.4$

$y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=$

$=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448$

$y(0.4)=3.005974448$

The Euler's Method is detailed in the following table.

<em>Point b:</em>

To find the general solution of $y'=3te^{-y}$ you need to:

Rewrite in the form of a first order separable ODE:

$e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt$

Integrate each side:

$\int \:e^ydy=e^y+C$

$\int \:3t\:dt=\frac{3t^2}{2}+C$

$e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}$

We know the initial condition y(0) = 3, we are going to use it to find the value of $C_{1}$

$e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3$

So we have:

$e^y=\frac{3t^2}{2}+e^3$

Solving for <em>y</em> we get:

$\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)$

<em>Point c:</em>

To compute the error in the approximations y(0.2), y(0.6), and y(1) you need to:

Find the values y(0.2), y(0.6), and y(1) using $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

$y(0.2)=\ln \left(\frac{3(0.2)^2}{2}+e^3\right)=3.002982771$

$y(0.6)=\ln \left(\frac{3(0.6)^2}{2}+e^3\right)=3.026529965$

$y(1)=\ln \left(\frac{3(1)^2}{2}+e^3\right)=3.072023504$

Next, where $y_{1}, y_{3}, \:and \:y_{5}$ are from the table.

$|y(0.2)-y_{1}|=|3.002982771-3|=0.002982771$

$|y(0.6)-y_{3}|=|3.026529965-3.017852169|=0.008677796$

$|y(1)-y_{5}|=|3.072023504-3.058523645|=0.013499859$

3 0

3 years ago