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makkiz [27]
3 years ago
10

Please help I don’t understand this and I can’t figure it out.

Mathematics
1 answer:
ryzh [129]3 years ago
7 0
So the groups of angles that make a line (like 130 and i or p a and t) should equal 180. however, the triangles should also equal 180. so in this case for the top row, we know that 130+j=180. since 180-130=50 (i did the inverse of addition, subtraction), j=50. now let’s say for that triangle that your top two angle measures were the 50 we just found and the 70 you have written just as an assumption but i have no clue what you were actually given. since triangles angles must equal 180, 50+70+that angle =180. 50+70=120, so if we do the inverse of 180-120=60, we find out that the bottom angle is 60 degrees.
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Find cos(2* angle ABC) 100 POINTS
Simora [160]

Answer:

3/5

Step-by-step explanation:

We need to use the trig identity that cos(2A) = cos²A - sin²A, where A is an angle. In this case, A is ∠ABC. Essentially, we want to find cos∠ABC and sin∠ABC to solve this problem.

Cosine is adjacent ÷ hypotenuse. Here, the adjacent side of ∠ABC is side BC, which is 4 units. The hypotenuse is 2√5. So, cos∠ABC = 4/2√5 = 2/√5.

Sine is opposite ÷ hypotenuse. Here, the opposite side of ∠ABC is side AC, which is 2 units. The hypotenuse is still 2√5. So sin∠ABC = 2/2√5 = 1/√5.

Now, cos²∠ABC = (cos∠ABC)² = (2/√5)² = 4/5.

sin²∠ABC = (sin∠ABC)² = (1/√5)² = 1/5

Then cos(2∠ABC) = 4/5 - 1/5 = 3/5.

8 0
3 years ago
Read 2 more answers
Match the cofactors to their corresponding entries in the matrix.
Masja [62]

Answer:

The co-factors and their values are shown in the table below.

Step-by-step explanation:

We are given the matrix, \begin{bmatrix}3&7&1\\7&1&-3\\8&5&1\end{bmatrix}.

It is required to match the co-factors with the corresponding values.

As, the co-factors are given by,

A_{c_{ij}} = (-1)^{i+j}(d), where the d= determinant of the matrix after removing the i- row and j- column.

So, we have,

1. A_{c_{11}}.

So, A_{c_{11}}=(-1)^{1+1}(1\times 1-5\times (-3))

i.e. A_{c_{11}}=(-1)^{2}(1+15)=16

2. A_{c_{13}}.

So, A_{c_{13}}=(-1)^{1+3}(7\times 5-1\times 8)

i.e. A_{c_{13}}=(-1)^{4}(35-8)=27

3. A_{c_{21}}.

So, A_{c_{21}}=(-1)^{2+1}(7\times 1-5\times 1)

i.e. A_{c_{21}}=(-1)^{3}(7-5)=-2

4. A_{c_{22}}.

So, A_{c_{22}}=(-1)^{2+2}(3\times 1-8\times 1)

i.e. A_{c_{22}}=(-1)^{4}(3-8)=-5

5. A_{c_{31}}.

So, A_{c_{31}}=(-1)^{3+1}(7\times (-3)-1\times 1)

i.e. A_{c_{31}}=(-1)^{4}(-21-1)=-22

Thus, we get,

Co-factor                 Value

A_{c_{11}}                              16

A_{c_{13}}                             27

A_{c_{21}}                             -2

A_{c_{22}}                            -5

A_{c_{31}}                           -22

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Answer:

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Step-by-step explanation:

Hope this helps

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Blababa [14]

Answer:

there is no solution for (0, 30°)

Step-by-step explanation:

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sin(t)^2 + cos(x)^2 - 2sin(x)cos(x) = 0

1 - sin(2x) = 0

sin(2x) = 1

2x = π/2 + 2πk -> k - any integer number

x = π/4 + πk

the smallest positive solution is π/4 = 45°. That means for range (0, 30°)

there is no solution

6 0
3 years ago
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