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Andru [333]
3 years ago
10

What is the equation in point slope form of the line that passes through the points (2,4) and has a slope of 3

Mathematics
2 answers:
strojnjashka [21]3 years ago
8 0

1) the equation in point slope form is

Y-Y1 = M(X -X1) , the point (x1,y1) is given in the exercise and is ( 2,4) and m is 3

then Y -4 =3(x-2)

Ronch [10]3 years ago
6 0

The formula for point slope is:

y-y1=m(x-x1)

m is your slope: 3; plug in 4 for y1 and 2 for x1

y - 4=3(x-2)

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Answer:

Step-by-step explanation:

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we know that

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∠ABD=∠BAD= ----> the angles of the base are equals

Find the measure of the vertex angle

∠ABD= ------> the sum of the internal angles of a triangle is equal to

Step 2

Find the measure of the angle  ∠CBD in the equilateral triangle

we know that

A equilateral triangle has three  equal sides and three equal angles

The measure of the internal angle in a equilateral triangle is

so

∠CBD=

Step 3

Find the measure of the angle ∠ABC

∠ABC=∠ABD+∠DBC

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the measure of the angle ∠ABC is

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Solving a Two-Step Matrix Equation<br> Solve the equation:
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Answer:

\boxed {x_{1} = 3}

\boxed {x_{2} = -4}

Step-by-step explanation:

Solve the following equation:

\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

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3x_{1} + 2x_{2} + 1 = 2

-Second equation:

5x_{1} + 5x_{2} + 2 = -3

-Choose one of the two following equations, which I choose the first one, then you solve for x_{1} by isolating

3x_{1} + 2x_{2} + 1 = 2

-Subtract 1 to both sides:

3x_{1} + 2x_{2} + 1 - 1 = 2 - 1

3x_{1} + 2x_{2} = 1

-Subtract 2x_{2} to both sides:

3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1

3x_{1} = -2x_{2} + 1

-Divide both sides by 3:

3x_{1} = -2x_{2} + 1

x_{1} = \frac{1}{3} (-2x_{2} + 1)

-Multiply -2x_{2} + 1 by \frac{1}{3}:

x_{1} = \frac{1}{3} (-2x_{2} + 1)

x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

-Substitute -\frac{2x_{2} + 1}{3} for x_{1} in the second equation, which is 5x_{1} + 5x_{2} + 2 = -3:

5x_{1} + 5x_{2} + 2 = -3

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

Multiply -\frac{2x_{2} + 1}{3} by 5:

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

-Combine like terms:

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

\frac{5}{3}x_{2} + \frac{11}{3} = -3

-Subtract \frac{11}{3} to both sides:

\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}

\frac{5}{3}x_{2} = -\frac{20}{3}

-Multiply both sides by \frac{5}{3}:

\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}

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x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

-Multiply -\frac{2}{3} and -4:

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

x_{1} = \frac{8 + 1}{3}

-Since both \frac{1}{3} and \frac{8}{3} have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

x_{1} = \frac{8 + 1}{3}

x_{1} = \frac{9}{3}

\boxed {x_{1} = 3}

5 0
3 years ago
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