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2 years ago

Use the diagram to answer the questions

Mathematics

1 answer:

harkovskaia [24]2 years ago

5 0

Answer:

sorry

Step-by-step explanation:

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A line passes through point A (14,21). A second point on the line has an x-value that is 125% of the x-value of point A and a y-

seropon [69]

Answer:

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

Step-by-step explanation:

According to the statement, let $A(x,y) = (14,21)$ and $B(x,y) = (1.25\cdot x_{A},0.75\cdot y_{A})$ . The equation of the line in point-slope form is defined by the following formula:

$y-y_{A} = m\cdot (x-x_{A})$ (1)

Where:

$x_{A}$ , $y_{A}$ - Coordinates of the point A, dimensionless.

$m$ - Slope, dimensionless.

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

In addition, the slope of the line is defined by:

$m = \frac{y_{B}-y_{A}}{x_{B}-x_{A}}$ (2)

If we know that $x_{A} = 14$ and $y_{A} = 21$ , then the equation of the line in point-slope form is:

$x_{B} = 1.25\cdot (14)$

$x_{B} = 17.5$

$y_{B} = 0.75\cdot (21)$

$y_{B} = 15.75$

From (2):

$m = \frac{15.75-21}{17.5-14}$

$m = -\frac{3}{2}$

By (1):

$y-21 = - \frac{3}{2}\cdot (x-14)$

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

5 0

3 years ago

Write in point slope form of the line that passes through the points (3,-1) and (6,-2)

skelet666 [1.2K]

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-2}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-2-(-1)}{6-3}\implies \cfrac{-2+1}{3}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-1)=-\cfrac{1}{3}(x-3)\implies y+1=-\cfrac{1}{3}(x-3)$

8 0

3 years ago

Solve the equation using the zero-product property. (2x − 8)(7x + 5) = 0 x = –2 or x = 7 x = 4 or x = x = 4 or x = x = –4 or x =

aliya0001 [1]

$(2x-8)(7x+5)=0\iff2x-8=0\ \vee\ 7x+5=0\\\\2x-8=0\qquad\text{add 8 to both sides}\\2x=8\qquad\text{divide both sides by 2}\\\boxed{x=4}\\\\7x+5=0\qquad\text{subtract 5 from both sides}\\7x=-5\qquad\text{divide both sides by 7}\\\boxed{x=-\dfrac{5}{7}}$