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ElenaW [278]
2 years ago
6

Someone knows this ??

Mathematics
1 answer:
Dima020 [189]2 years ago
8 0

Answer: A (1)

Explanation:

a(11) means column 1, row 1. Therefore, the answer would be the top left value.

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See the picture below please help me and thank you.
svetoff [14.1K]

Answer: -1.13

Step-by-step explanation:

6 0
2 years ago
Which of the following is not equal to sin⁡(-230°)?
xenn [34]

From trigonometry we know that:

if sin(\theta)=sin(\alpha)

then, \theta=n\pi+(-1)^n\alpha (where n is an integer)

This can be rewritten in degrees as:

\theta=n(180^{\circ})+(-1)^n\alpha.............(Equation 1)

Now, in our case, \alpha=-230^{\circ}

Therefore, (Equation 1) can be written as:

\theta=n(180^{\circ})+(-1)^n(-230^{\circ})..........(Equation 2)

Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of \theta that match.

For n=2, (Equation 2) gives us: \theta=2\times 180^{\circ}+(-1)^2(-230^{\circ})=360^{\circ}-230^{\circ}=130^{\circ}.

Thus, sin(230^{\circ})=sin(130^{\circ})

Now, we know that: -sin(-50^{\circ})=sin(50^{\circ})

Let n=-1, then:

\theta=(-1)\times 180^{\circ}+(-1)^{-1}(-230^{\circ})=-180^{\circ}+230^{\circ}=50^{\circ}

Thus, sin(-230^{\circ})=-sin(-50^{\circ})

Likewise, sin(-230^{\circ})=sin(50^{\circ})

Only the last option sin(-50^{\circ}) will never match sin(-230^{\circ}) because no integral value of n will ever give \theta=-50^{\circ}

Thus the last option is the correct option.

3 0
2 years ago
Read 2 more answers
Find the equation of the line that passes through the point (3,14) and is parallel to y=2/3x -6
Ugo [173]
Y=mx+b
m=slope

paralell ines have same slope
y=2/3x-6
slope is 2/3

the equation of aline that passes through (x1,y1) and has a slope of m is
y-y1=m(x-x1)
given
(3,14) and slope is 2/3
y-14=2/3(x-3)
y-14=2/3x-2
y=2/3x+12
4 0
3 years ago
Read 2 more answers
Solve step by step<br>2(p+8)=26
romanna [79]

Answer:

<h2>p = 5</h2>

Step-by-step explanation:

2(p+8)=26\qquad\text{divide both sides by 2}\\\\\dfrac{2\!\!\!\!\diagup^1(p+8)}{2\!\!\!\!\diagup_1}=\dfrac{26\!\!\!\!\!\diagup^{13}}{2\!\!\!\!\diagup_1}\\\\p+8=13\qquad\text{subtract 8 from both sides}\\\\p=5

8 0
3 years ago
Read 2 more answers
Abby used the law of cosines for TriangleKMN to solve for k.
torisob [31]

Answer:

  m∠K = 37° and n = 31

Step-by-step explanation:

A lot of math is about matching patterns. Here, the two patterns we want to match are different versions of the same Law of Cosines relation:

  • a² = b² +c² -2bc·cos(A)
  • k² = 31² +53² -2·31·53·cos(37°)

<h3>Comparison</h3>

Comparing the two equations, we note these correspondences:

  • a = k
  • b = 31
  • c = 53
  • A = 37°

Comparing these values to the given information, we see that ...

  • KN = c = 53 . . . . . . . . . . matching values 53
  • NM = a = k . . . . . . . . . . . matching values k
  • KM = b = n = 31 . . . . . . . matching values 31
  • ∠K = ∠A = 37° . . . . . . . matching side/angle names

Abby apparently knew that ∠K = 37° and n = 31.

__

<em>Additional comment</em>

Side and angle naming for the Law of Sines and the Law of Cosines are as follows. The vertices of the triangle are labeled with single upper-case letters. The side opposite is labeled with the same lower-case letter, or with the two vertices at either end.

Vertex and angle K are opposite side k, also called side NM in this triangle.

5 0
1 year ago
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