All categories

3 years ago

I really need math help, it’s slopes

Mathematics

1 answer:

zalisa [80]3 years ago

6 0

Same LOL first tru the ones with the rise and run showed then try the others remember y=mx+b and your rules

You might be interested in

14 1/5 - 5 5/6 as a fraction

melomori [17]

Answer:

$\frac{251}{30}$ or 8 $\frac{11}{30}$

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Damon measured his bedroom and the length measured 5.6 meters. Given that I meter = 3.28 feet, what is the length of Damon's roo

Elina [12.6K]

Answer:

D) 18.37 feet

Step-by-step explanation:

5.6 x 3.28 = 18.37 feet

6 0

3 years ago

What is the scale factor of 3

Wittaler [7]

A scale factor of 3 means that the new shape is three times the size of the original. To calculate the Scale Factor, we use the following: You can get the 'big' and 'small' from the corresponding sides on the figures. So just get your # and multiply it by 3

7 0

3 years ago

PLS HELP I WILL GIVE 15 POINTS OR MORE PLS NO SPAMMING PLS

Blizzard [7]

B,D,F / I’m pretty sure I did this not to long ago

3 0

3 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

3 years ago