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3 years ago

Rewrite the following parametric equations in rectangular form. x=e^3t y=e^-t

1 answer:

5 0

To solve this problem you must apply the proccedure shown below:
1. You have the following parametric equations given in the problem above:
<span> x=e^3t
y=e^-t</span>
2. Therefore, you must solve fot et in the second equation, as below:
y=e^-t
y=1/e^t
e^t=1/y
3.Substitute into the first equation:
x=e^3t
x=(e^t)^3
x=(1/y)^3
x=1/y^3 (y>0)
The answer is: x=1/y^3 (y>0)

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Select the correct answer.

Eva8 [605]

Answer:

Equation C: 5/8x + 2..5 = 3/8x + 1.5 + 1/4x

Step-by-step explanation:

Let us go through each of the equations one by one.

Equation A: $2.3y+2+3.1y=4.3y+1.6+1.1y+0.4$

This simplifies to

$5.4y+2=5.4y+2$

which has infinitely many solutions.

Equation B: $32x+25-21x=10x$

this simplifies to

$x+25=0$

$x=-25$

which is exactly one solution.

Equation C: $\frac{5}{8} x+2.5=\frac{3}{8}x+1.5+\frac{1}{4} x$

this simplifies to

$\frac{5}{8} x+2.5=\frac{5}{8} +1.5$

$2.5=1.5$

There's no way this is going to be possible; this equation has no solutions.

Equation D: $\frac{1}{3} x+\frac{1}{7} x=\frac{3}{7} x$

this gives

$\frac{1}{3} =\frac{2}{7} y$

$y=\frac{7}{6}$

which is exactly one solution.

Thus we see that it is equation C that has no solutions.

3 0

3 years ago

Determine whether the value is a discrete random variable, continuous random variable, or not a random variable.

Viktor [21]

Discrete random variable as we count the number of free dash throw attempts and we dont measure them (continuous)

7 0

3 years ago

Mademuasel [1]

Answer:

I only

Step-by-step explanaition:

The domain (x) can not repeat in order for it to be a function.

3 0

2 years ago

Read 2 more answers

WILL MARK U AS BRAINLIEST PLZ HELP

Radda [10]

Answer:

112 if you don't include the triangle sides as they are not necessary, but 124 if you include one triangle and 136 if you include both triangle sides

5 0

2 years ago

Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose P(V) = 0.17

ArbitrLikvidat [17]

Answer: 0.82

Step-by-step explanation:

The probability of the computer not containing neither a virus nor a worm is expressed as P( $V^{C}$ ∩ $W^{C}$ ) , where P( $V^{C}$ ) is the probability that the event V doesn't happen and P( $W^{C}$ ) is the probability that the event W doesn't happen.

P( $V^{C}$ )= 1-P(V) = 1-0.17 = 0.83

P( $W^{C}$ )=1-P(W) = 1-0.05 = 0.95

Since $V^{C}$ and $W^{C}$ aren't mutually exclusive events, then:

P( $V^{C}$ ∪ $W^{C}$ ) = P( $V^{C}$ ) + P( $W^{C}$ ) - P( $V^{C}$ ∩ $W^{C}$ )

Isolating the probability that interests us:

P( $V^{C}$ ∩ $W^{C}$ )= P( $V^{C}$ ) + P( $W^{C}$ ) - P( $V^{C}$ ∪ $W^{C}$ )

Where P( $V^{C}$ ∪ $W^{C}$ ) = 1 - 0.04 = 0.96

Finally:

P( $V^{C}$ ∩ $W^{C}$ ) = 0.83 + 0.95 - 0.96 = 0.82

5 0

2 years ago