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nasty-shy [4]
4 years ago
12

A system of equations is shown below. Which of the following statements describes the graph of this system of equations in the (

x, y) coordinate plane?
3y − 5x = 15
6y − 10x = 30

Select one:
A. Two parallel lines with positive slope
B. Two parallel lines with negative slope
C. A single line with positive slope
D. A single line with negative slope
Mathematics
2 answers:
katovenus [111]4 years ago
7 0

Answer:

C

Step-by-step explanation:

3y - 5x = 15   ... A

y = 5/3 * X +15  .... slope 3/5 , y intercept 15

6y - 10x =30

(6y - 10x) /2 = 30 / 2

3y - 5x = 15 ... B

y = 5/3 * x + 15 ..... slope 3/5 , y intercept 15

∴ Answer is C

bezimeni [28]4 years ago
5 0

Answer:

Option A. Two parallel lines with positive slope.

Step-by-step explanation:

Thinking process:

An equation of a straight line is written as:

ay = mx + c

where a = is the coefficient of y

          m =     the gradient of the line

          c = y -intercept

Examining the two equations:

3y − 5x = 15          ...1

6y − 10x = 30       ...2

The equations can be rearranged as follows:

3y = 5x + 15  

6y = 10 x + 30 or 2 (3y = 5 x + 15)

As seen from the expression, the equation (2) is twice the equation (1)

In addition, the equations are equal if (2) is reduced to lowest terms.

The gradients will be equal.

The equations are parellel (equal gradient) with a positive slope (5 and 10 respectively)

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AlladinOne [14]

Answer:

Yes

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
2. Calculate an expression for dy/dx and d2y/dx2 in terms of t if the parametric pair is given as tan(x) = e^at and e^y = 1 + e^
Ber [7]

I assume a is a constant. If tan(x) = exp(at) (where exp(x) means eˣ), then differentiating both sides with respect to t gives

sec²(x) dx/dt = a exp(at)

Recall that

sec²(x) = 1 + tan²(x)

Then we have

(1 + tan²(x)) dx/dt = a exp(at)

(1 + exp(2at)) dx/dt = a exp(at)

dx/dt = a exp(at) / (1 + exp(2at))

If exp(y) = 1 + exp(2at), then differentiating with respect to t yields

exp(y) dy/dt = 2a exp(2at)

(1 + exp(2at)) dy/dt = 2a exp(2at)

dy/dt = 2a exp(2at) / (1 + exp(2at))

By the chain rule,

dy/dx = dy/dt • dt/dx = (dy/dt) / (dx/dt)

Then the first derivative is

dy/dx = (2a exp(2at) / (1 + exp(2at))) / (a exp(at) / (1 + exp(2at))

dy/dx = (2a exp(2at)) / (a exp(at))

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Since dy/dx is a function of t, if we differentiate dy/dx with respect to x, we have to use the chain rule again. Suppose we write

dy/dx = f(t)

By the chain rule, the derivative is

d²y/dx² = df/dx

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d²y/dx² = (df/dt) / (dx/dt)

d²y/dx² = 2a exp(at) / (a exp(at) / (1 + exp(2at)))

d²y/dx² = 2 (1 + exp(2at))

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3 years ago
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wariber [46]
C because i took the test
3 0
3 years ago
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Choose the point-slope form of the equation below that represents the line that passes through the point (6, -3) with a slope of
Firlakuza [10]
The point-slope form is
y = mx + b
m, the slope, is 1/2.
To find b, we plug in the point (6,-3) to the equation:
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Therefore, the point-slope equation is:
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