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Natalka [10]
2 years ago
10

Solve, then write, each equation in slope intercept form. ​

Mathematics
1 answer:
djyliett [7]2 years ago
4 0

Answer:

1) (3,5)

2) (4,-7)

3)  (6,4)

4) (-4,-3)

Step-by-step explanation:

1)

Subtract  y  from both sides of the equation.

x  =  8  −  y

x  −  y  =  −2

Replace all occurrences of x  in x  −  y  =  −2  with  8  −  y.

x  =  8  −  y

(8  −  y)  −  y  =  −2

Subtract y from  −y.

x  =  8  − y

8  −  2y  =  −2

Solve for  y  in the second equation.

x  =  8  −  y  

y  = 5

Replace all occurrences of  y  in  x  =  8  −  y  with  5.

x  =  8  −(5)

y  =  5

Simplify  8  −(5).

x  =  3

y  =  5

Slope = (3,5)

2)

Multiply each term by  2  and simplify.

2x+y=12

y+6=−x−4

Move all terms containing variables to the left side of the equation.

2x+y=1x+2

y+6=−4

Move all terms not containing a variable to the right side of the equation.

2x+y=1x+2

y=−10

Multiply each equation by the value that makes the coefficients of  x  opposite.

2x+y=1(−2)⋅

(x+2y)=(−2)(−10)

Simplify.

2x+y=1−2x−4

y=20

Add the two equations together to eliminate  x  from the system.

2x+y=1+−2x−4

y=20−3

y=21

Divide each term by  −3  and simplify.

y  = −7

Substitute the value found for  y  into one of the original equations, then solve for  x.

x  =  4

slope (4,-7)

3)

Subtract  y  from both sides of the equation.

2x−y=8x−y=2

Multiply each equation by the value that makes the coefficients of  y  opposite

2x−y=8(−1)⋅

(x−y)=(−1)(2)

Simplify. 2x−y=8

−x+y=−2

Add the two equations together to eliminate  y  from the system.

2x−y=8  +  −x+y=−2x=6

Substitute the value found for  x  into one of the original equations, then solve for  y.

y=4

slope (6,4)

4)

Simplify 2(x+1).

y−3=2x+2

y=−3(x+5)

Simplify  −3(x+5)

y−3=2x+2

y=−3x−15

Move all terms containing variables to the left side of the equation.

−2x+y−3=2

y=−3x−15

Add  3x  to both sides of the equation.

−2x+y−3=2

y+3x=−15

Move all terms not containing a variable to the right side of the equation.

−2x+y=5

y+3x=−15

Reorder the polynomial.

−2x+y=5

3x  +  y=−15

Multiply each equation by the value that makes the coefficients of  y  opposite.

−2x+y=5(−1)⋅

(3x+y)=(−1)(−15)

Simplify.

−2x+y=5

−3x−y=15

Add the two equations together to eliminate  y  from the system.

−2x+y=5  +  −3x−y=15−5x=20

Divide each term by  −5  and simplify.

x=−4

Substitute the value found for  x  into one of the original equations, then solve for y.

y=−3

slope (-4,-3)

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<h3>Properties of Logarithms</h3>

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

  • Product Rule for Logarithms - log_{b}(a*c)=log_{b}a+log_{b}c
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The exercise asks the potential solutions for  log_4x+log_4(x+6)=2. In this expression you can apply the Product Rule for Logarithms.

                                  log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0

Now you should solve the quadratic equation.

 

 Δ=b^2-4ac=36-4*1*(-16)=36+64=100. Thus, x will be x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}. Then:

x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8

The potential solutions  are 2 and -8.

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Lina20 [59]

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Two triangles having two pairs of congruent sides and a pair of congruent angles do not necessarily meet the SAS congruence criterion, therefore Angie is incorrect.

<h3 /><h3>What is congruency?</h3>

The Side-Angle-Side Congruence Theorem (SAS) defines two triangles to be congruent to each other if the included angle and two sides of one is congruent to the included angle and corresponding two sides of the other triangle.

An included angle is found between two sides that are under consideration.

See image attached below that demonstrates two triangles that are congruent by the SAS Congruence Theorem.

Thus, two triangles having two pairs of corresponding sides and one pair of corresponding angles that are congruent to each other is not enough justification for proving that the two triangles are congruent based on the SAS Congruence Theorem.

The one pair of corresponding angles that are congruent MUST be "INCLUDED ANGLES".

Therefore, based on the SAS congruence criterion, the statement that best describes Angie's statement is:

Two triangles having two pairs of congruent sides and a pair of congruent angles do not necessarily meet the SAS congruence criterion, therefore Angie is incorrect.

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Gnesinka [82]

The values of a and b are 1/4 and 1/5 respectively

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The above function is a horizontal hyperbola.

A horizontal hyperbola that passes through the origin is represented as:

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