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3 years ago

*giving brainiest to the first correct answer. Write the equation of the line graphed.

Mathematics

2 answers:

Oduvanchick [21]3 years ago

7 0

Answer:

x = 2

Step-by-step explanation:

When you are given a straight line whether it's horizontal or vertical on a graph and is told to write it as an equation,

it can be a horizontal line, y = the number the line is on from the y-axis

or a vertical line, x = the number the line is on from the x-axis.

In this case you have a vertical line so the equation would be

x = the number the line is on from the x-axis, which is 2 so your final answer would be

x = 2

AURORKA [14]3 years ago

6 0

Answer:

The equation of the line is x = 2

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In a right triangle one leg measures a quarter of the other, how much do its acute angles measure? with procedure please

KATRIN_1 [288]

Answer:

The answer to your question is 14° and 76°

Step-by-step explanation:

Data

leg 1 = x

leg 2 = x/4

angle 1 = ?

angle 2 = ?

Process

To solve problem use trigonometric functions. We must use the trigonometric function that relates both legs (tangent).

tangent Ф = Opposite side / Adjacent side

-Substitution

tan Ф = (x/4) / x

-Simplification

tan Ф = x / 4x

tan Ф = 1/4

- Find Ф

Ф = tan⁻¹(1/4) = 14°

- Find the other acute angle

The sum of the internal angles in a triangles equals 180°

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Step-by-step explanation:

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3 years ago

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Integration of ∫(cos3x+3sinx)dx

Murljashka [212]

Answer:

$\boxed{\pink{\tt I = \dfrac{1}{3}sin(3x) - 3cos(x) + C}}$

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

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Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
Now , Rewrite using du and u .

$\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I = \dfrac{1}{3}sin(3x) - 3cos(x) + C }}}}}$

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3 years ago

A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.

kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is y=3x−13

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is 13 , the slope of a line perpendicular to line BC is −3 . The equation of altitude 3 is y=−3x−1

The orthocenter is the point where all three altitudes intersect.

x=2

y=3x−13

y=−3x−1

Use substitution to solve the first two equations y=3(2)−13=−7

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we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1

−7=−6−1

−7=−7 it works.

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