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3 years ago

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On the blueprint of a house, 36 millimeters represents 7 meters. The length of the living room is 27 millimeters on the blueprin

t. What is the actual length of the living room?

1 answer:

S_A_V [24]3 years ago

7 0

36/27=1.3
7x1.3=9.1
The actual length is 9.1meters

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each classroom in willett hall will hold 34 students.There are 4 floors with 6 classroom on each floor. How many students can be

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4 floors with 6 class rooms each = 6*4 = 24 total class rooms.

24 class rooms x 34 students in each one = 816 total students.

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The measures of two pairs of corresponding angles of two triangles are 24 degrees and 55 degrees. Explain why the two triangles

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Answer:

Are the two triangles below similar?

Triangles ABC and DEF are shown. Angle A measure 30 degrees, angle C measure 65 degrees, side AC measures 14, side AB measures

Yes, because there are two pairs of congruent corresponding angles

No, because there are not two pairs of congruent corresponding angles

Yes, because the corresponding sides are proportional

No, because the corresponding sides are not proportional

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2 years ago

Find the third side in simplest radical form:<br> 25

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Answer:

<h3> $\boxed{ \bold{24}}$ </h3>

Step-by-step explanation:

$\mathsf{given}$

$\mathsf{hypotenuse(h) = 25}$

$\sf{perpendicular (p) = 7}$

$\sf{base(b) = }$ ?

Now, Using Pythagoras theorem

$\sf{{h}^{2} = {p}^{2} + {b}^{2} }$

plug the values

⇒ $\sf{ {25}^{2} = {7}^{2} + {b}^{2} }$

Evaluate the power

⇒ $\sf{625 = 49 + {b}^{2} }$

Swap the sides of the equation

⇒ $\sf{49 + {b}^{2} = 625}$

Move constant to right hand side and change it's sign

⇒ $\sf{ {b}^{2} = 625 - 49}$

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⇒ $\sf{ {b}^{2} = 576}$

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⇒ $\sf{b = 24}$

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3 years ago

What prevent is modeled below

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Answer:

B 35%

Step-by-step explanation:

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3 years ago

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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