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Lemur [1.5K]
3 years ago
7

A rectangular wall measures 1620cm by 68cm. Estimate the area of the wall.

Mathematics
1 answer:
mixer [17]3 years ago
5 0
Length×width≈area
1620×68≈110160cm²
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Solve: 2x+3y=1<br> y=-2x+11
Sauron [17]

Answer:

(8, -5); x=8; y=-5

Step-by-step explanation:

i'll be solving via substitution

2x+3y=1; y=-2x+11

2x + 3(-2x +11) = 1

2x -6x +33 = 1

-4x = -32

x = 8

2(8) + 3y = 1

16 + 3y = 1

3y = -15

y = -5

(8, -5)

Check:

-5 = -2(8) + 11

-5 = -16 + 11

-5 = -5

6 0
3 years ago
Find the missing angle measurements. Screenshot included.
zheka24 [161]
One of the missing angles is 45degrees 64+71=135 then subtract from 180 since it is a triangle. Then 64 are vertical angles with the one across from it which stays the same. The last angle would be 64+56=120 and 180-120=60 which would be the last angle. Hope this Helps... Juan
3 0
3 years ago
Daniel wants to create a banner for his construction business to hang at the house he just built. Daniel will use a single sign
Nina [5.8K]
The solution for this problem is just simple. It tells you to find the area in square feet. Thus, convert the units to feet first. The conversion you need to know are: 12 in = 1 ft and 3 ft = 1 yard.

Width = 42 in * 1 ft/12 in = 3.5 ft
Length = 10 yards * 3 ft/1 yard = 30 ft

Area = Length*Width = 30*3.5
<em>Area = 105 ft²</em>
3 0
3 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
Math Help PLZ!!!!!!!!!
VMariaS [17]

Answer: C

<u>Step-by-step explanation:</u>

 f(x) =   2x + 1  ⇒       2x + 1

<u>- g(x) = -(x² - 7) </u>⇒  <u>-x²     + 7 </u>

              (f-g)(x) = -x² + 2x + 8

4 0
3 years ago
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