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qwelly [4]
3 years ago
7

Write the equation of the line that includes the point (1,5) and has a slope of 3 in standard form.

Mathematics
2 answers:
Ymorist [56]3 years ago
7 0

Answer:

-3x + y = 2

Step-by-step explanation:

The slope is 3. So the coefficient of x is 3.

when x = 1, y = 5.

So 3*1 + c = 5

So c is 2.

So y = 3x + 2

Therefore, taking 3x to the LHS, we get,

-3x + y = 2

Hope it helps ;)

Kay [80]3 years ago
3 0

For this case we have that the equation of the point-slope form is given by:

(y-y_ {0}) = m (x-x_ {0})

Where:

m: It's the slope

(x_ {0}, y_{0}): It is a point through which the line passes

We have to:

(x_ {0}, y_{0}) = (1,5)\\m = 3

Then, the equation is:

y-5 = 3 (x-1)\\y-5 = 3x-3\\-5 + 3 = 3x-y\\-2 = 3x-y

We multiply by "-" on both sides of the equation:

-3x + y = 2

Answer:

Option B

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Teresa should build 7.5 inches model.

Step-by-step explanation:

Consider the provided information.

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First find how many 61's are there in 305.

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2 years ago
Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1
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N 1)
a²<span> − 9a + 14 = 0 
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 9a+20.25)=-14+20.25

Rewrite as perfect squares

(a-4.5)²=6.25--------> (a-4.5)=(+/-)√6.25

a1=4.5+√6.25-----> a1=7

a2=4.5-√6.25-----> a2=2

the solution problem N 1 is the pair {7, 2}


N 2) 

a²<span> + 9a + 14 = 0
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Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

(a+4.5)²=6.25--------> (a+4.5)=(+/-)√6.25

a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3) 

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

the solution problem N 3 is the pair {2, -5}


N 4)

a²<span> + 5a − 14 = 0
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Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 5a+6.25) =14+6.25

Rewrite as perfect squares

(a+2.5)² =20.25-------> (a+2.5)=(+/-)√20.25

a1=-2.5+√20.25-----> a1=2

a2=-2.5-√20.25-----> a2=-7

the solution problem N 4 is the pair {2, -7}


N 5) 

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

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(a² − 5a+6.25)=14+6.25

Rewrite as perfect squares

(a-2.5)²=2025--------> (a-2.5)=(+/-)√20.25

a1=2.5+√20.25-----> a1=7

a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

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