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3 years ago

10

i am a regular polygon with all obtuse angles. I have the smallest number of side of any polygon with obtuse angles. How may sid

e do I have?

1 answer:

andreyandreev [35.5K]3 years ago

6 0

You would have four sides

You might be interested in

Are the following figures similar?

Tcecarenko [31]

No the corresponding angles are not congruent, because the angle measures on the smaller figure are 90, 90, 137, and 43, while the larger figure has angle measures of 90, 90, 136, 44. that is why the following figures are not congruent.

3 0

3 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

4 years ago

In a set of 25 aluminum castings, four castings are defective (D), and the remaining twenty-one are good (G). A quality control

lianna [129]

Answer:

The sample space for selecting the group to test contains <u>2,300</u> elementary events.

Step-by-step explanation:

There are a total of <em>N</em> = 25 aluminum castings.

Of these 25 aluminum castings, <em>n</em>₁ = 4 castings are defective (D) and <em>n</em>₂ = 21 are good (G).

It is provided that a quality control inspector randomly selects three of the twenty-five castings without replacement to test.

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!(n-k)!}$

Compute the number of samples that are possible as follows:

${25\choose 3}=\frac{25!}{3!\times (25-3)!}$

$=\frac{25\times 24\times 23\times 22!}{3!\times 22!}\\\\=\frac{25\times 24\times 23}{3\times 2\times 1}\\\\=2300$

The sample space for selecting the group to test contains <u>2,300</u> elementary events.

6 0

3 years ago

Suppose that Y varies directly with X andY equals 12 when X equals -2 what is X when Y equals -6

VLD [36.1K]

Answer:

$x=1$

Step-by-step explanation:

A direct variation equation has the following format:

$y=kx$

Where k is a constant.

We know that y is 12 when x is -2. Thus, substitute:

$12=-2k$

Divide both sides by -2:

$k=-6$

So, our direct variation equations is:

$y=-6x$

To find what x is when y equals -6, substitute in -6 for y:

$-6=-6x$

Divide both sides by -6:

$x=1$

So, our answer is 1 :)

4 0

4 years ago

Select all the ordered pairs that make this statement true?

Ierofanga [76]

Answer: (24,-9)

(0,9)

(4,6)

Step-by-step explanation:

3x + 4y = 36

Using (3,-2) will be:

= 3(3) + 4(-2)

= 9 - 8 = 1

Using (1,7) will be

= 3(1) + 4(7)

= 3 + 28 = 31

Using (0,0) equals to 0

Using (24,-9) will be:

= 3(24) +4( -9)

= 72 - 36 = 36

Using (0,9) will be:

= 3(0) + 4(9)

= 0 + 36 = 36

Using (4,6) will be:

3(4) + 4(6)

= 12 + 24

= 36

Using (-12,18) will be:

= 3(-12) + 4(-18)

= -36 - 72

= -108

The correct options are (24,-9), (0,9 and (4,6)

5 0

3 years ago