All categories

3 years ago

Which of the point-slope equations below are correct for the line that passes through the points (6,6) and (3,1)?

1 answer:

4 0

(6,6)(3,1)
slope(m) = (1 - 6) / (3 - 6) = -5/-3 = 5/3

y - y1 = m(x - x1)...using (6,6)
y - 6 = 5/3(x - 6) <== here is one

y - y1 = m(x - x1)...using (3,1)
y - 1 = 5/3(x - 3) <== here is the other one

You might be interested in

Determine which graph represents a reflection across the x-axis of f(x) = 3(1.5).

Ghella [55]

Step-by-step explanation:

fill in the missing fraction parts

4 0

1 year ago

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

In triangle XYZ right angled at Y. Find the length of the hypotenuse if the length of the other two sides is 1.6 cm and 6.3 cm.

const2013 [10]

Answer:

6.5 cm

Step-by-step explanation:

From Pythagoras theorem

C^2= a^2 + b^2

Where a, b are the sides of the triangle

C= hypotenus sides of the triangle

two sides are given as 1.6 cm and 6.3cm respectively.

C^2= 1.6^2 + 6.3^2

C^2= 2.56 + 39.69

C^2 = 42.25

C= √42.25

C=6.5 cm

Hence, the length of the hypotenuse is 6.5 cm

7 0

2 years ago

Solve the given equation. ( 1/5 )( 4/7 )x = 32/490 (x is and exponent) x= _____

Vikki [24]

X= 4/7 because it would be

4/35x=16/245

which would be 28x=16

simplified it would be 4/7

5 0

3 years ago

of the 273 students in a school, 95 volunteered to work the book sale. About what percent of the students volunteered?

Whitepunk [10]

28.7% to b exact
I hope this helped you

6 0

3 years ago