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alina1380 [7]
4 years ago
11

(7)(8)(9) Geometry Check Ans 7) s = 15 8) y = 64 x = 99 9) 96

Mathematics
1 answer:
Katyanochek1 [597]4 years ago
6 0
Problem 7: Correct
Problem 8: Correct
Problem 9: Correct

The steps are below if you are curious

===========================================================================================

Problem 7

S = 180*(n-2)
2340 = 180*(n-2)
2340/180 = n-2
13 = n-2
n-2 = 13
n = 13+2
n = 15

I'm using n in place of lowercase s, but the idea is the same. If anything, it is better to use n for the number of sides since S already stands for the sum of the interior angles. I'm not sure why your teacher decided to swap things like that.

===========================================================================================

Problem 8

First find y
y+116 = 180
y+116-116 = 180-116
y = 64

which is then used to find x. The quadrilateral angles add up to 180*(n-2) = 180*(4-2) = 360 degrees
Add up the 4 angles, set the sum equal to 360, solve for x

x+y+125+72 = 360
x+64+125+72 = 360 ... substitution (plug in y = 64)
x+261 = 360
x+261-261 = 360-261
x = 99

===========================================================================================

Problem 9

With any polygon, the sum of the exterior angles is always 360 degrees

The first two exterior angles add to 264. The missing exterior angle is x
x+264 = 360
x+264-264 = 360-264
x = 96

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Gnesinka [82]
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Which of the following
Archy [21]

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Thus meaning 36 is the odd number out. I hope this helps!

6 0
4 years ago
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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
3 years ago
Find the inequalitie for d thank youuu <br><br><br> -7d+8&gt;29
shutvik [7]

Answer:

d<3

Step-by-step explanation:

-7d+8>29

First lets remove the inequality  for know

-7d+8=29

then remember to re add the inequality to the equation

-7d>21

since you divide by negative you switch

d<3

that is your answer

3 0
3 years ago
Change 7/5 to a whole number
Harlamova29_29 [7]
Write it as a mixed fraction 7/5 = 1 2/5
5 0
4 years ago
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