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2 years ago

A line passes through the point (-3,-3) and has a slope of 1/2 what is the equation of the line ?

Mathematics

1 answer:

Bezzdna [24]2 years ago

7 0

Answer:

y = 1/2x -1.5

Step-by-step explanation:

The equation of a line is typically written as y = mx + b.

m is the slope of the line, while b is the y-intercept. y is the y-coordinate and x is the x-coordinate.

Since it indicated that the line has a slope of 1/2, we can substitute the m in the equation with 1/2.

y = 1/2x + b

In order to find the intercept of the line, we use the equation of a line to substitute the y-coordinate and x-coordinate of (-3,-3) to discover the y-intercept of b.

-3 = 1/2(-3) + b

One half of -3 is -1.5.

-3 = -1.5 + b

Add -1.5 to both sides of the equation.

-3 = -1.5 + b

+1.5 +1.5

-1.5 = b

Since we found the y-intercept, we can now place it into our equation.

y = 1/2x -1.5 and that's the answer!

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Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

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Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

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By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

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3 0

3 years ago

Use the diagram to determine which statement is true

hichkok12 [17]

A. Area of ABCD - Area of DGA = Area of DEFG
s^2 - 1/2bh = s^2
(5)^2 - 1/2(4)(3) = (3)^2
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25 - 24 = 9
1 not equal to 9

B. Area of ABCD - Area of GHIA = Area of DGA
s^2 - s^2 = 1/2bh
(5)^2 - (4)^2 = 1/2(4)(3)
25 - 16 = 1/2(12)
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C. Area of ABCD + Area of DGA = Area of GHIA
s^2 + 1/2bh = s^2
(5)^2 + 1/2(4)(3) = (4)^2
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25 + 6 = 16
31 not equal to 16

D. Area of DEFG + Area of GHIA = Area of ABCD
s^2 + s^2 = s^2
(3)^2 + (4)^2 = (5)^2
9 + 16 = 25
25 = 25

The answer is D.

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