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3 years ago

Create a system of equations that are parallel to the y-axis.

Mathematics

1 answer:

pishuonlain [190]3 years ago

4 0

Answer:

the y-axis equation that are parallel is x-axis

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Do,2 of Y is: (5, 4) (6,4) (3/2,1)

kotegsom [21]

Answer:

(6,4)

Step-by-step explanation:

This question asks you to dilate by a factor of 2, and from the origin. Because it is from the origin, we can simply multiply the coordinates of Y by 2

3*2=6

2*2=4

so the result is 6,4

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4 years ago

A large order of cheesy fries costs $7. How much would it costs to buy three orders of cheesy fries?

Maru [420]

Answer:

21 dollars

Step-by-step explanation:

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3 years ago

I need help with these problems please

DanielleElmas [232]

Answer:

20=5 because if a =(x+20)° then as my solution b=5

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4 years ago

How do you solve this linear equation 3x+2=x+2x+4 ?

Svetach [21]

3x+2=x+2x+4
-2 -2
3x=x+2x+4-2
3x=3x+4-2
-3x -3x
x= 4-2
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7 0

3 years ago

Read 2 more answers

Help with numer 5 please. thank you

Alex17521 [72]

Answer:

See Below.

Step-by-step explanation:

We are given that:

$\displaystyle I = I_0 e^{-kt}$

Where I₀ and k are constants.

And we want to prove that:

$\displaystyle \frac{dI}{dt}+kI=0$

From the original equation, take the derivative of both sides with respect to t. Hence:

$\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right]$

Differentiate. Since I₀ is a constant:

$\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)$

Using the chain rule:

$\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right) = -kI_0e^{-kt}$

We have:

$\displaystyle \frac{dI}{dt}+kI=0$

Substitute:

$\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0$

Distribute and simplify:

$\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0$

Hence proven.

4 0

3 years ago