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MrRissso [65]
3 years ago
15

First Identify the Angle relationship, then solve for x.

Mathematics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

Corresponding Angles Theorem

x = -5

Step-by-step explanation:

Since they are corresponding angles, they are equal to each other.

x + 105 = 100

x = -5

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Point P is the center of the circle below. What is the exact area
Ber [7]

Answer:

192π unit²

Step-by-step explanation:

Given that :

Radius = 24

Area of circle = πr²

Area = π*24²

Area = 576π in²

The Shaded area is 120°

Entire Circumference = 360°

Hence, shaded area = 120°/ 360° = 1/ 3 of the area

1/3 * 576π in²

= 192π unit²

5 0
2 years ago
Find the measure of the missing angle<br><br> Please help me, I’ll give you points
Leno4ka [110]

Answer:

a = 28°

Step-by-step explanation:

62°+a° = 90°

a = 90 - 62

a = 28°

5 0
2 years ago
Read 2 more answers
What is the slope of -y=-x+6
Katarina [22]

<u>Answer:</u>

  • m = 1
  • b = -6

<u>Step-by-step explanation:</u>

We know that:

  • -y = -x + 6
  • y = mx + b

First, we need to change the 'y' sign.

  • => -y = -x + 6
  • => y = x - 6

Now, let's compare both of the equations to find slope.

  • => (y = x - 6) = (y = mx + b)

We can see that the slope is 1 and the y-intercept is -6.

<u>Conclusion:</u>

Therefore:

  • m = 1
  • b = -6

Hoped this helped.

4 0
2 years ago
WILLL GIVE POINTS
Eddi Din [679]

Answer:

(1,1) (0,1) (2,-2) (3,2)

Step-by-step explanation:

So with dilation you have to mulitply thE numbers by the one said in the problem which Was 1/3. So after doing that I got those coordinates.

5 0
3 years ago
Please help me with this.
OLga [1]

Swapping rows alters the sign of the determinant:

\begin{vmatrix} x & y & z \\ -8 & 2 & -12 \\ u & v & w \end{vmatrix} = - \begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix}

Multiplying a single row by a scalar scales the determinant by the same amount:

\begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix} = -2 \begin{vmatrix} x & y & z \\ u & v & w \\ 4 & -1 & 6 \end{vmatrix}

Then

\begin{vmatrix} x & y & z \\ -8 & 2 & -12 \\ u & v & w \end{vmatrix} = -(-2) \begin{vmatrix} x & y & z \\ u & v & w \\ 4 & -1 & 6 \end{vmatrix} = 2\times(-6) = \boxed{-12}

8 0
2 years ago
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