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3 years ago

Determine if the lines are parallel, perpendicular or neither y=-3/2x-4 and y=2/3x-7

Mathematics

1 answer:

aliya0001 [1]3 years ago

4 0

Answer:

They are perpendicular

Step-by-step explanation:

Perpendicular lines have slopes that are negated versions of the other or positive versions of the slope of the other. The slopes are also flipped.

Parallel lines have the same slope.

Neither is when the lines do not follow either of these definitions.

The slope of your two lines are -3/2 and 2/3. These slopes are flipped and the opposite negations of each other. This means the lines are perpendicular.

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Find an equation for the line that passes through the points (5.-5) and (-4,-2)

ANEK [815]

Answer:

$\large\boxed{y=-\dfrac{1}{3}x-\dfrac{10}{3}\to x+3y=-10}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (5, -5) and (-4, -2).

Substiute:

$m=\dfrac{-2-(-5)}{-4-5}=\dfrac{-2+5}{-9}=\dfrac{3}{-9}=-\dfrac{1}{3}$

Put the value of the slope and the coordinates of the point (5, -5) to the equation of a line:

$-5=-\dfrac{1}{3}(5)+b$

$-5=-\dfrac{5}{3}+b$ add 5/3 to both sides

$-\dfrac{15}{3}+\dfrac{5}{3}=b\\\\-\dfrac{10}{3}=b\to b=-\dfrac{10}{3}$

Finally we have the equation of a line in the slope-intercept form:

$y=-\dfrac{1}{3}x-\dfrac{10}{3}$

Convert to the standard form (Ax + By = C):

$y=-\dfrac{1}{3}x-\dfrac{10}{3}$ multiply both sides by 3

$3y=-x-10$ add x to both sides

$x+3y=-10$

4 0

3 years ago

Help how many questions did you answer correctly

Lady bird [3.3K]

Answer:

450

Step-by-step explanation:

you divide 90 by 20 and times in by 100

4 0

3 years ago

Read 2 more answers

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

Camera 2nd has to cover the maximum angle, i.e. $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$