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4 years ago

Find the perimeter of a square with an area of 361 square feet

Mathematics

2 answers:

alina1380 [7]4 years ago

8 0

19 feet x 4 = 76 feet

lisabon 2012 [21]4 years ago

4 0

Answer: 76

Step-by-step explanation:

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Find the midpoint of the line segment whose endpoints are (-2, 5) and (6, -9). (4, -4) (2, -2) (1, -4)

Rus_ich [418]

Answer:

(2,-2)

Step-by-step explanation:

The midpoint of the line segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ has coordinates

$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

If the endpoints of the line segment are (-2,5) and (6,-9), then the midpoint has the coordinates

$\left(\dfrac{-2+6}{2},\dfrac{5+(-9)}{2}\right)\\ \\\left(\dfrac{4}{2},\dfrac{-4}{2}\right)\\ \\(2,-2)$

8 0

3 years ago

Question 15

Hoochie [10]

Answer:

10% chance or 1/10

Step-by-step explanation:

5 out of the 100 cards are free lunches simplified it would be

1/10

8 0

4 years ago

Read 2 more answers

Solve the following equation for x. x2 + 8x + 7 = 0 A. x = -1; x = -7 B. x = 1; x = -7 C. x = 1; x = 7 D. x = -1; x = 7

BabaBlast [244]

So the right answer is of option A.

Look at the attached picture

Hope it will be helpful to you..

5 0

3 years ago

Read 2 more answers

The line has been partitioned into three angles. Is there a triangle with these three angle measures?

nexus9112 [7]

La curva de 120 grados

3 0

3 years ago

Section 5.2 Problem 21:

Fittoniya [83]

Answer:

$y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$ (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation $am^2+bm+c=0$ where the values of $m$ are the roots:

$y''+4y'+10y=0\\\\m^2+4m+10=0\\\\m^2+4m+10-6=0-6\\\\m^2+4m+4=-6\\\\(m+2)^2=-6\\\\m+2=\pm\sqrt{6}i\\\\m=-2\pm\sqrt{6}i$

Since the values of $m$ are complex conjugate roots, then the general solution is $y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]$ where $m=\alpha\pm\beta i$ .

Thus, the general solution for our given differential equation is $y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]$ .

To account for both initial conditions, take the derivative of $y(x)$ , thus, $y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]$

Now, we can create our system of equations given our initial conditions:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3$

$y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2$

We then solve the system of equations, which becomes easy since we already know that $C_1=3$ :

$-2C_1+\sqrt{6}C_2=-2\\\\-2(3)+\sqrt{6}C_2=-2\\\\-6+\sqrt{6}C_2=-2\\\\\sqrt{6}C_2=4\\\\C_2=\frac{4}{\sqrt{6}}\\ \\C_2=\frac{4\sqrt{6}}{6}\\ \\C_2=\frac{2\sqrt{6}}{3}$

Thus, our final solution is:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$

7 0

2 years ago