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Andrew [12]
2 years ago
10

Please help it’s a test and I only have 50 minutes left

Mathematics
1 answer:
viktelen [127]2 years ago
5 0

Answer:

60

Step-by-step explanation:

X is equal to 60 degrees

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Find m∠U.<br> Write your answer as an integer or as a decimal rounded to the nearest tenth.
olga2289 [7]

Answer:

Answer:

50.2

Step-by-step explanation:

ut =  \sqrt{ {6}^{2}  +  {5}^{2} }   \\  \sqrt{36 + 25}  \\  \sqrt{61 }  \\ using \: cosine \: rule \\ cos \: u =  \frac{ {5}^{2} + 61 - 36 }{2 \times 5 \times  \sqrt{61} }  \\  \\  =  \frac{25 + 61 - 36}{78}  \\  \\  = \frac{50}{78 }  \\  \cos \: u = 0.64 \\ m∠U. = 50.2

7 0
2 years ago
6c - (3 – 3c) = 24<br> Solve equation step by step
kap26 [50]

Hi there! :)

\large\boxed{c = 3}

6c - (3 - 3c) = 24

Distribute the negative sign with terms inside the parenthesis;

6c - (3) - (-3c) = 24

6c - 3 + 3c = 24

Combine like terms:

9c - 3 = 24

Add 3 to both sides:

9c = 27

Divide both sides by 9:

c = 3.

7 0
3 years ago
Read 2 more answers
The smaller triange was dilated to form the larger triangle. What is the value of x?
natka813 [3]

Answer:

x = 14

Step-by-step explanation:

As we can see that the bottom side is equivalent to 3 of the smallest triangle and the side of the larger triangle is equivalent to the 9

This represents that it is dilated by 3 factor

So

= 3 × 3

= 9

Now the other side i.e. 5 of the smallest triangle also need to be dilated by a 3 factor

So the new side is 15

Also the largest triangle is x + 1

So,

15 = x + 1

x = 14

4 0
3 years ago
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
3 years ago
Can you turn these inequalities into y=mx+b form? <br> y&gt; 3x-6<br><br><br> 4x-y&lt;7
kaheart [24]
I am not sure but wouldn't it be y=3x-6 and y=4x-7
8 0
3 years ago
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