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3 years ago

14

The area of a rectangle increased by seven is the same as for less than twice the area of the rectangle. Right and solve an equa

tion to determine the area of the rectangle

1 answer:

never [62]3 years ago

7 0

Answer:

11 Square Units

Step-by-step explanation:

Let A be the area of the Rectangle

First statement: The area of the rectangle is increased by seven, we have :

A+7

Second Statement: Four less than twice the area of the rectangle

Twice the Area=2A

Four less = 2A-4

Therefore:

2A-4=A+7

Collecting like terms

2A-A =7+4

A=11

The area of the rectangle is 11 Square Units.

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35. V112 + V63<br> How do I solve this please

konstantin123 [22]

Answer: 175V

Step-by-step explanation:

V112+V63=112v+63V

Add them together because the terms are like terms. Therfore, 112V+63V=175V

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3 years ago

You buy a rental property for $180,000. Assuming that you could sell the property for $250,000 at the end of 6 years, what is yo

docker41 [41]

Answer:

15.542%

Step-by-step explanation:

For uneven cash flows such as those in this problem, there is no formula for "internal rate of return" (IRR). It must be computed graphically or iteratively. Spreadsheets and financial calculators are equipped to do this calculation. Attached is the result of the calculation done by a graphing calculator.

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7 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

What fraction of a batch of trail mix consists of peanuts

Luden [163]

2 and 2/3
This will be tha answer I tjink

5 0

3 years ago

Someone please solve and explain part a(i)

Crank

Step-by-step explanation:

the way I understand the description :

C is below B. they are on a kind of straight hill, and there is a straight "road" going up from C to B.

then, at B there is an antenna or other firm of mast going straight up.

therefore, this is not a right-angled triangle with 90 degrees at B (as it would be, if this would be in a flat plane).

but because it goes downhill from B to C the angle is 105 degree.

(a)(i)

now, imagine, there would be a horizontal plane either at B or at C. AB would have a true 90 degree angle with this plane. so, what is the angle of CB with this plane ?

this angle is the "excess" of the 90 degrees, as CB angles down from the horizontal plane at B, or angles up with the same angle from the horizontal plane at C.

what is the "excess" of 105 degrees vs. the standard 90 degrees ? 105 - 90 = 15 degrees.

(a)(ii)

the extended Pythagoras for not right-angled triangles :

c² = a² + b² - 2ab×cos(B)

B being the angle opposite of the Hypotenuse c.

so, we have

c² = 15² + 10² -2×15×10×cos(105) = 225 + 100 - 300×cos(105) =

= 325 - 300×cos(105) = 402.6457135...

c = 20.06603383... ≈ 20 m

4 0

3 years ago