All categories

3 years ago

Please help me please

Mathematics

2 answers:

myrzilka [38]3 years ago

8 0

Answer:

B. y = -4/3x + 22

Step-by-step explanation:

Parallel lines have the same slope.

<u>Use the given point (6, 14) and point-slope form:</u>

y - 14 = -4/3(x - 6)
y = -4/3x + 8 + 14
y = -4/3x + 22

Correct choice is B

Dennis_Churaev [7]3 years ago

7 0

Answer:

Solution given:

y=-4/3×x-1.......(1)

comparing above equation with y=mx+c

slope of line 1 is

m=-4/3

since the another line is parallel

slope is equal

m=m1=-4/3.

since it passes through (6,14)

we have

equation of a line is:

(y-y1)=m1(x-x1)

(y-14)=-4/3(x-6)

y=-4/3×x+4/3×6+14

y=-4/3×x+8+14

y=-4/3×x+22.

your option is b.y=-4/3×x+22.

You might be interested in

Which relation does not represent a function

Nastasia [14]

Answer:

For example, if given a graph, you could use the vertical line test; if a vertical line intersects the graph more than once, then the relation that the graph represents is not a function.

Hope this helped

3 0

3 years ago

The x-coordinate of point P is positive, and the y-coordinate of point P is positive. In which quadrant is point P?

mestny [16]

The answer is Quadrant 1

8 0

3 years ago

Read 2 more answers

Solve each quadratic equation. Show your work 15. x2 + 3x = 10

OLEGan [10]

X² + 3x = 10

Convert to standard form.
x² + 3x - 10 = 0

factor x² = x * x
factor -10 = 5 * -2

(x + 5) (x - 2) = 0
x(x -2) + 5(x-2) = 0
x² - 2x + 5x - 10 = 0
x² + 3x - 10 = 0

x + 5 = 0   x² + 3x = 10
x = -5 -5² + 3(-5) = 10
25 - 15 = 10
10 = 10

x - 2 = 0 x² + 3x = 10
x = 2 2² + 3(2) = 10
   4 + 6 = 10
   10 = 10

4 0

3 years ago

Read 2 more answers

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is: