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3 years ago

Find the value of X (Gotta choose at least 2) x = 52 x = 2 x = 3 x = 33

Mathematics

2 answers:

Ad libitum [116K]3 years ago

8 0

Answer:

Before this problem gets an answer, can you define "gotta." Does this work mean isosceles?

Step-by-step explanation:

lana [24]3 years ago

4 0

Answer:

x = 3

Step-by-step explanation:

To solve this problem, we need to remember the exterior angle theorem. This theorem tells us that: "The exterior angle of a triangle is equal to the sum of the two opposite interior angles"

In the picture, we would have that the angle ∠QRT =∠RTS + ∠TSR

Substituting the values from the figure we get:

$45x=25x + 57+x\\45x=26x+57\\45x-26x=57\\19x=57\\x=3$

Thus, the value for x is 3

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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

3 years ago

Dilate (9, 6) using a scale factor of 1/3

KATRIN_1 [288]

The answer is (3,2)

To get this answer we either...
A) Multiply both coordinates by 1/3
or
B) Divide each coordinate by 3

9*(1/3) = 9/3 = 3
6*(1/3) = 6/3 = 2

6 0

4 years ago

Many credit card companies charge a compound interest rate of 1.8% per month on a credit card balance. Nelson owes $850 on a cre

Finger [1]

The sequence will be

$850, 850 \times (1.018)^1, 850 \times (1.018)^2, 850 \times (1.018)^3, 850 \times (1.018)^4, ...$