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2 years ago

14

To rent a boat, you pay for an initial fee plus. John rented a boat for 3 hours and paid $41. Jessica rented a boat for 6 hours

and paid $77.

1 answer:

8090 [49]2 years ago

6 0

Answer:

The intial fee is $5

The equation to represent this is: y=12x+5

12 is the cost per hour x=hours y=total cost

Step-by-step explanation:

77-41=36

36/3=12

12*6=72

77-72=5

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A line has a slope of -6 over 7. What is the slope of the line parallel to it? And what is the slope of the line perpendicular t

olga_2 [115]

The slope of the parallel line is -6/7
the slope of the perpendicular line is 7/6
the slope of the line = -6/7
the gradient of two parallel lines are equal
the product of the gradient of two perpendicular lines is -1
: the gradient m1*m2 = -1
m2= -1(-6/7)
m2= 7/6

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2 years ago

-9 · (-8) =<br> O 72<br> 00<br> o<br> 00<br> O-72

Masja [62]

Answer:

72

Step-by-step explanation:

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2 years ago

The points (-10, 0) and (-5, 3) fall on a particular line. What is it’s equation in slope-intercept form?

yaroslaw [1]

Answer:

y = 3/5x + 6

Step-by-step explanation:

When given two points, it is convenient to start with the 2-point form of the equation of a line, then rearrange it to the form needed for the problem. For points (x1, y1) and (x2, y2), the equation is ...

y = (y2 -y1)/(x2 -x1)·(x -x1) +y1

Filling in the given point coordinates, this is ...

y = (3 -0)/(-5 -(-10))·(x -(-10)) + 0

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2 years ago

Find the equation of the parabola that has zeros of x = –2 and x = 3 and a y–intercept of (0,–30).

Anit [1.1K]

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2 years ago

Find limit as x approaches 2 from the left of the quotient of the absolute value of the quantity x minus 2, and the quantity x m

Molodets [167]

Answer:

$\lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1$ .

Step-by-step explanation:

We want to find $\lim_{x \to 2^-} \frac{|x-2|}{x-2}$ .

By definition:

$|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:$

Since we want to find the Left Hand Limit, we use f(x)=-(x-2)

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}$ .

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} (-1)$ .

The limit of a constant is the constant.

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1$ .

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2 years ago