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arsen [322]
3 years ago
6

Draw a line representing the "rise" and a line representing the "run" of the line. State the slope of the line in simplest form.

Mathematics
1 answer:
Paul [167]3 years ago
8 0
The slope is 3/4
you just count how many it goes up before it goes over to a whole unit
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What is the equation of the line that is parallel to the line 5x + 2y = 12 and passes through the point (−2, 4)? 
astra-53 [7]
What is the equation of the line that is parallel to the line 5x + 2y = 12 and passes through the point (−2, 4)?

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In Mr. Chang's class, Jack found out that his walking rate is 2.5 meters per second. That is, Jack
sergejj [24]

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Please help! If the current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms, what is the voltage?
Leona [35]

The value of the voltage is 18 - j volts if the current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms option (c) is correct.

<h3>What is a complex number?</h3>

It is defined as the number which can be written as x+iy where x is the real number or real part of the complex number and y is the imaginary part of the complex number and i is the iota which is nothing but a square root of -1.

We have:

The current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms.

As we know,

V = IR

V = (3 + j2)(4 - j3)

\rm V=\left(3\cdot \:4-2\left(-3\right)\right)+\left(3\left(-3\right)+2\cdot \:4\right)j

j² = -1 (complex part)

After simplification:

V = 18 - j  volts

Thus, the value of the voltage is 18 - j volts if the current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms option (c) is correct.

Learn more about the complex number here:

brainly.com/question/10251853

#SPJ1

6 0
2 years ago
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = t, y
DochEvi [55]

Answer:

Step-by-step explanation:

At the point (0, 1,0) t = 0

Find the tangent vector:

\frac{dx}{dt}= 1

\frac{dy}{dt}= -4e^{-4t}

\frac{dz}{dt}=5-5t^4

The tangent vector for all points \vec v(t) is

\vec v(t) = \hat {i}-4e^{-4t}\hat{j}+(5-5t^4)\hat{k}

\rightarrow \vec v (0)= \hat{i}-4\hat{j}

The vector equation of the tangent line is

(x,y,z) = (0,1,0)+s(\hat{i}-4\hat{j})

The parametric equation for this line are

x= s

y=1-4s

z=0

7 0
4 years ago
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