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11 months ago

32. What is the perimeter of a regular octagon with a side length of 7 units?

Mathematics

2 answers:

Maurinko [17]11 months ago

7 0

Answer:

56 units

Step-by-step explanation:

the perimeter of a rectangular octagon with sides of length a

is equal to: 8 × a

Then

the perimeter of a regular octagon with a side length of 7 units

is equal to: 8 × 7 = 56 units

katovenus [111]11 months ago

4 0

Answer:

Step-by-step explanation:

solution

Here

P=8a=8·7=56

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Angelina_Jolie [31]

Answer:

Delta does not have a relative extreme in the equation, so it is variable y It can not be true

Step-by-step explanation:

$s = 5x + 3y$

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The rest of the equation that has no substituted delta is not true, so the prime variables have relative extremes.

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Find the EXACT value of sec (-120 degrees)

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2 years ago

Read 2 more answers

4x + 9 = 2x + 15 .. I need help

valkas [14]

Answer:

x = 3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Define Equation

4x + 9 = 2x + 15

Step 2: Solve for x

Subtract 2x on both sides: 2x + 9 = 15
Subtract 9 on both sides: 2x = 6
Divide 2 on both sides: x = 3

Step 3: Check

Plug in x into the original equation to verify it's a solution.

Substitute in x: 4(3) + 9 = 2(3) + 15
Multiply: 12 + 9 = 6 + 15
Add: 21 = 21

Here we see that 21 does indeed equal 21.

∴ x = 3 is the solution to the equation.

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2 years ago

consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cros

Bezzdna [24]

Answer:

$q(x,y,z)=16x^{2}-5y^{2}-5z^{2}$

Step-by-step explanation:

The given quadratic form is of the form

$q(x,y,z)=ax^2+by^2+dxy+exz+fyz$ .

Where $a=11,b=-1,c=-4,d=-16,e=8,f=-4$ .Every quadratic form of this kind can be written as

$q(x,y,z)={\bf x}^{T}A{\bf x}=ax^2+by^2+cz^2+dxy+exz+fyz=\left(\begin{array}{ccc}x&y&z\end{array}\right) \left(\begin{array}{ccc}a&\frac{1}{2} d&\frac{1}{2} e\\\frac{1}{2} d&b&\frac{1}{2} f\\\frac{1}{2} e&\frac{1}{2} f&c\end{array}\right) \left(\begin{array}{c}x&y&z\end{array}\right)$

Observe that $A$ is a symmetric matrix. So $A$ is orthogonally diagonalizable, that is to say, $D=Q^{T}AQ$ where $Q$ is an orthogonal matrix and $D$ is a diagonal matrix.

In our case we have:

$A=\left(\begin{array}{ccc}11&(\frac{1}{2})(-16) &(\frac{1}{2}) (8)\\(\frac{1}{2}) (-16)&(-1)&(\frac{1}{2}) (-4)\\(\frac{1}{2}) (8)&(\frac{1}{2}) (-4)&(-4)\end{array}\right)=\left(\begin{array}{ccc}11&-8 &4\\-8&-1&-2\\4&-2&-4\end{array}\right)$

The eigenvalues of $A$ are $\lambda_{1}=16,\lambda_{2}=-5,\lambda_{3}=-5$ .

Every symmetric matriz is orthogonally diagonalizable. Applying the process of diagonalization by an orthogonal matrix we have that:

$Q=\left(\begin{array}{ccc}\frac{4}{\sqrt{21}}&-\frac{1}{\sqrt{17}}&\frac{8}{\sqrt{357}}\\\frac{-2}{\sqrt{21}}&0&\sqrt{\frac{17}{21}}\\\frac{1}{\sqrt{21}}&\frac{4}{\sqrt{17}}&\frac{2}{\sqrt{357}}\end{array}\right)$

$D=\left(\begin{array}{ccc}16&0&0\\0&-5&0\\0&0&-5\end{array}\right)$

Now, we have to do the change of variables ${\bf x}=Q{\bf y}$ to obtain

$q({\bf x})={\bf x}^{T}A{\bf x}=(Q{\bf y})^{T}AQ{\bf y}={\bf y}^{T}Q^{T}AQ{\bf y}={\bf y}^{T}D{\bf y}=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\lambda_{3}y_{3}^{2}=16y_{1}^{2}-5y_{2}^{2}-5y_{3}^2$

Which can be written as:

$q(x,y,z)=16x^{2}-5y^{2}-5z^{2}$

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3 years ago

Help please & show work

Natasha2012 [34]

Answer:

Infected = itself + itself

Healthy = 200 - infected

y axis = weeks

x axis = # of people

use green for healthy and red for infected

and mark going by weeks

green:

(1, 199)

(2, 198)

(3, 196)

(4, 192)

(5, 184)

(6, 168)

(7, 136)

(8, 72)

(9, -56)

red:

(1, 1)

(2, 2)

(3, 4)

(4, 8)

(5, 16)

(6, 32)

(7, 64)

(8, 128)

(9, 256)

(10, 512)

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2 years ago