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Oxana [17]
2 years ago
13

Write the equation of the line perpendicular to F(x)=5 that goes through (-1, -3)

Mathematics
1 answer:
Ainat [17]2 years ago
3 0

Answer:

x = -1

Step-by-step explanation:

y = 5 is a horizontal line

It's perpendicular would be a vertical line of the form:

x = a

Since it passes through (-1,-3)

a = -1

Hence line is: x = -1

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2 years ago
The equation of a line is y=3. Write an equation in slope intercept form of a line parallel to y=3 that passes through (0,6).
Dimas [21]

Answer:

y = 6

Step-by-step explanation:

We have to consider the equation of a line ( in slope - intercept form ) that is parallel to y = 3, and passes through the point ( 0, 6 ). Now y = 3 can be plotted as a horizontal line that passes through the point ( 0, 3 ). If we want a line that is parallel to this, it must be horizontal as well;

Let us consider the second point now. This line must pass through the point ( 0, 6 ). We can conclude that the line must be 1. Horizontal, and 2. Pass through point ( 0, 6 );

Equation - y = 6

7 0
3 years ago
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Lubov Fominskaja [6]

Answer:

See Below.

Step-by-step explanation:

We are given the isosceles triangle ΔABC. By the definition of isosceles triangles, this means that ∠ABC = ∠ACB.

Segments BO and CO bisects ∠ABC and ∠ACB.

And we want to prove that ΔBOC is an isosceles triangle.

Since BO and CO are the angle bisectors of ∠ABC and ∠ACB, respectively, it means that ∠ABO = ∠CBO and ∠ACO = ∠BCO.

And since ∠ABC = ∠ACB, this implies that:

∠ABO = ∠CBO =∠ACO = ∠BCO.

This is shown in the figure as each angle having only one tick mark, meaning that they are congruent.

So, we know that:

\angle ABC=\angle ACB

∠ABC is the sum of the angles ∠ABO and ∠CBO. Likewise, ∠ACB is the sum of the angles ∠ACO and ∠BCO. Hence:

\angle ABO+\angle CBO =\angle ACO+\angle BCO

Since ∠ABO =∠ACO, by substitution:

\angle ABO+\angle CBO =\angle ABO+\angle BCO

Subtracting ∠ABO from both sides produces:

\angle CBO=\angle BCO

So, we've proven that the two angles are congruent, thereby proving that ΔBOC is indeed an isosceles triangle.

7 0
3 years ago
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Convert 86400cm² into meters​
vredina [299]
864000cm squared to meters squared would be 8.64
6 0
3 years ago
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Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.
FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

\rho = \sqrt{x^{2}+y^{2}+z^{2}}

\phi = cos^{-1}\frac{z}{\rho}

For angle θ:

  • If x > 0 and y > 0: \theta = tan^{-1}\frac{y}{x};
  • If x < 0: \theta = \pi + tan^{-1}\frac{y}{x};
  • If x > 0 and y < 0: \theta = 2\pi + tan^{-1}\frac{y}{x};

Calculating:

a) (4,2,-4)

\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}} = 6

\phi = cos^{-1}(\frac{-4}{6})

\phi = cos^{-1}(\frac{-2}{3})

For θ, choose 1st option:

\theta = tan^{-1}(\frac{2}{4})

\theta = tan^{-1}(\frac{1}{2})

b) (0,8,15)

\rho = \sqrt{0^{2}+8^{2}+(15)^{2}} = 17

\phi = cos^{-1}(\frac{15}{17})

\theta = tan^{-1}\frac{y}{x}

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = \frac{\pi}{2}

c) (√2,1,1)

\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}} = 2

\phi = cos^{-1}(\frac{1}{2})

\phi = \frac{\pi}{3}

\theta = tan^{-1}\frac{1}{\sqrt{2} }

d) (−2√3,−2,3)

\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}} = 5

\phi = cos^{-1}(\frac{3}{5})

Since x < 0, use 2nd option:

\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }

\theta = \pi + \frac{\pi}{6}

\theta = \frac{7\pi}{6}

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

r=\sqrt{x^{2}+y^{2}}

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

r=\sqrt{4^{2}+2^{2}} = \sqrt{20}

\theta = tan^{-1}\frac{1}{2}

z = -4

b) (0, 8, 15)

r=\sqrt{0^{2}+8^{2}} = 8

\theta = \frac{\pi}{2}

z = 15

c) (√2,1,1)

r=\sqrt{(\sqrt{2} )^{2}+1^{2}} = \sqrt{3}

\theta = \frac{\pi}{3}

z = 1

d) (−2√3,−2,3)

r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}} = 4

\theta = \frac{7\pi}{6}

z = 3

5 0
3 years ago
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