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2 years ago

What is RS? Put the answer below. RS =

Mathematics

2 answers:

katovenus [111]2 years ago

7 0

Rs is 66, Sixty six, according to diagram

gulaghasi [49]2 years ago

7 0

The person above me is right it’s 66

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Pollsters are concerned about declining levels of cooperation among persons contacted in surveys. A pollster contacts 95 people

Marrrta [24]

Answer:

Step-by-step explanation:

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8 0

2 years ago

Can somebody please help me with this equation

Simora [160]

The answer is c your welcome

5 0

2 years ago

Is there an ordered pair that is a solution to BOTH of

Arte-miy333 [17]

Answer: The ordered pair is (2,3)

Step-by-step explanation:

To solve this problem, we can use substitution. With substitution, we substitute the y in each problem with the equations. In this case, x + 1 can substitute the y in y = -x + 5. Here is what the problem looks like now:

x + 1 = -x + 5

The reason we used substitution is because we assume that these linear equations are equal. We must prove it, though. We can use algebra.

Move the variables to one side and the numerical values to another:

x + x = 5 - 1.

Simplify.

2x = 4

Simplify.

x = 2.

Now, we got our x value, we can substitute it into our problem:

y = 2 + 1

y + -2 + 5

2 + 1 = - 2 + 5 = 3

Our x is 2, and our y is 3. The ordered pair that is a solution is (2,3).

6 0

3 years ago

Pls Help

vodomira [7]

First solve the triangle:
75° + 75° + x = 180°
150° + x = 180°
x = 180° - 150°
Therefore x = 30°
Then, use the geometrical property of vertically opposite angles.
Therefore. x° = 30°

3 0

2 years ago

Read 2 more answers

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

2 years ago