A. Graph B because the slope is at a constant rate
B. Graph A because since it is slowly opened the rise of the slope starts of small then rises more
C. Graph C since it’s opened a lot the beginning and closed at the end so the rise would start high then go lower as time goes by

8 0

4 years ago

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Answer before 1:00 i will mark you brainliest

Alecsey [184]

Answer:

the circumference of the circle= 36pi units

Step-by-step explanation:

5 0

3 years ago

How do I solve this algebra two btw

pychu [463]

$\bf \begin{cases} x^2+y^2=400\\ \boxed{y}=x-28 \end{cases}\qquad \stackrel{\textit{substituting on the 1st equation}}{x^2+\left( \boxed{x-28} \right)^2=400} \\\\\\ x^2+(x^2-56x+28^2)=400\implies x^2+x^2-56x+28^2=400 \\\\\\ 2x^2-56x+784=400\implies 2x^2-56x+384=0 \\\\\\ 2(x^2-28x+192)=0\implies x^2-28x+192=0 \\\\\\ (x-16)(x-12)=0\implies x = \begin{cases} 16\\ 12 \end{cases}$

now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then

$\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill$

7 0

3 years ago

Write the equation of the line that passes through (3, 4) and (2, −1) in slope-intercept form. y = 3x − 7 y = 3x − 5 y = 5x − 11

babymother [125]

Find the slope of the line first using $\frac{y2 - y1}{x2 - x1}$
(3, 4), (2, -1)
$\frac{-1 - 4}{2 - 3}$
$\frac{-5}{-1}$ = 5

Now use y - y1 = m(x - x1) to make an equation in slope-intercept form. Plug the numbers into the formula.

y - (-1) = 5(x - 2)
y + 1 = 5x - 10
y = 5x - 11

The equation of the line in slope-intercept form would be y = 5x - 11.

3 0

3 years ago

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