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Paul [167]
2 years ago
15

Triangles 3 and 4 are identical. They are joined in the poster to form a square. Construct a math argument to show why triangles

3 and 4 are isosceles right angles.

Mathematics
1 answer:
Bond [772]2 years ago
7 0

Triangles 3 and 4 are isosceles right angles because they have same length.

<h3>How to explain the triangle?</h3>

It should be noted be noted that the square has all right angles equal. The largest angle of the triangle is the right angle m.

Here, the triangles are isosceles right angles as both the legs and the acute angles are congruent.

Learn more about triangles on:

brainly.com/question/17335144

#SPJ1

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Answer !! ( PLEASE SHOW WORK FOR NUMBER 5.) ( ZOOM IN TO SEE BETTER)
deff fn [24]

Answer:

12+2x

Step-by-step explanation:

2(10)+2(x-4)

20+2x-8

12+2x

5 0
3 years ago
-4(-6 - 2n) = 30 + 8n
kati45 [8]

Answer:

n=6/4

Step-by-step explanation:

-4(-6-2n)=30+8n

24+12n=30+8n

-24       =-24

12n=6+8n

-8n =   -8n

4n=6

n=6/4

8 0
2 years ago
There are two square roots of 25. Which best explains why there is only one answer for the radius of the button?
Brrunno [24]
5 times5 is the one 5 times itself is 25 hents 25
4 0
3 years ago
What is the area of this regular pentagon that has been divided into five congruent triangles?
Len [333]

Answer:

120 cm^2.

Step-by-step explanation:

The area of one triangle = 1/2 * base * height

= 1/2 * 8 * 6

= 1/2 * 48

= 24 cm^2

So area of the pentagon = 5 * 24

= 120 cm^2.

7 0
2 years ago
In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
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