Mathematics, 27.06.2019 14:50 adrian08022

Answer asap i need to graduate this weekthe coordinates (4, -1). what17. a line has a point with the coordinates (-3,-2) and a point with the coordis the slope of the line? a. 1-osla mala18. a line has a point with the coordinates (0,6) and a point with the coordithe slope of this line? with the coordinates (2,9). what isglow

Answer asap i need to graduate this weekthe coordinates (4, -1). what17. a line has a point with the

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Answer from: VamPL

Question 1:

For this case we have that by definition, the slope of a line is given by:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

Where:

$(x_ {1}, y_{1}) and (x_ {2}, y_{2})$ are two points through which the line passes.

$(x_ {1}, y_{1}) = (- 3, -2)\\(x_ {2}, y_{2}) = (4, -1)$

Substituting in the equation:

$m = \frac {-1 - (- 2)} {4 - (- 3)} = \frac {-1 + 2} {4 + 3} = \frac {1} {7}$

Option C

Question 2:

For this case we have that by definition, the slope of a line is given by:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

Where:

$(x_ {1}, y_{1}) and (x_ {2}, y_{2})$ are two points through which the line passes.

$(x_ {1}, y_{1}) = (0,6)\\(x_ {2}, y_{2}) = (2,9)$

Substituting in the equation: $m = \frac {9-6} {2-0} = \frac {3} {2}$

Option A

Question 3:

For this case we have that by definition, the slope of a line is given by the following formula:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

Where:

$(x_ {1}, y_{1}) and (x_ {2}, y_{2})$ are two points through which the line passes.

We have as data:

$m = \frac {7} {8}\\(x_ {1}, y_{1}) = (- 2,1)$

Substituting in the formula:

$\frac {7} {8} = \frac {y_ {2} -1} {x_ {2} - (- 2)}\\\frac {7} {8} = \frac {y_ {2} -1} {x_ {2} +2}$

We substitute each of the points and see if the equality is met:

Point A: (6,8)

$\frac {7} {8} = \frac {8-1} {6 + 2}\\\frac {7} {8} = \frac {7} {8}$

Equality is met.

Option A

Question 4:

For this case we have that by definition, the slope of a line is given by the following formula:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

Where:

$(x_ {1}, y_{1}) and (x_ {2}, y_{2})$ are two points through which the line passes.

We have as data:

$m = \frac {1} {6}\\(x_ {1}, y_{1}) = (0, -3)$

Substituting in the formula:

$\frac {1} {6} = \frac {y_ {2} - (- 3)} {x_ {2} -0}\\\frac {1} {6} = \frac {y_ {2} +3} {x_ {2} -0}$

We substitute each of the points and see if the equality is met:

Point A: (-3,0)

$\frac {1} {6} = \frac {0 + 3} {- 3-0}\\\frac {1} {6} = \frac {3} {- 3}$

It is not fulfilled!

Point B: (6, -2)

$\frac {1} {6} = \frac {-2 + 3} {6-0}\\\frac {1} {6} = \frac {1} {6}$

Equality is met!

Option B

Answer from: Quest

i understand 100% 10/10 hit or miss?

step-by-step explanation:

Who is and what is this person i'm just being weird nobody answer this

Answer from: Quest

starting with:

5 + x - 12 = 2x - 7

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)

5 - 12 = x - 7 (combine and cancel x's from both sides)

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)

0 = x (add numbers together on both sides)

so in order for the equation to be true, x must be 0.

trying x = -0.5

5 + x - 12 = 2x - 7 (original equation)

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)

5 + -0.5 - 12 = -1 - 7 (did multiplication)

-7.5 = -8 (did addition on both sides)

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.

trying x = 0

5 + x - 12 = 2x - 7 (original equation)

5 + 0 - 12 = 2(0) - 7 (plug in value of x)

5 + 0 - 12 = 0 - 7 (did multiplication)

-7 = -7 (did addition on both sides)

since -7 equals -7, that demonstrates that the value of 0 for x is correct.

trying x = 1

5 + x - 12 = 2x - 7 (original equation)

5 + 1 - 12 = 2(1) - 7 (plug in value of x)

5 + 1 - 12 = 2 - 7 (did multiplication)

-6 = -5 (did addition on both sides)

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

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