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tia_tia [17]
3 years ago
9

Is y=4x+3 and 4x+y=3 parallel or perpendicular or neither?

Mathematics
1 answer:
Sedbober [7]3 years ago
3 0
Neither of the equations are parallel or perpendicular
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Please help I will give brainliest
Oksi-84 [34.3K]

Answer:

-273c

Step-by-step explanation:

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The solution to the system is (6,2) Select all the equations that could be the other equation in the system. PLZ HELPPPPPP there
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9514 1404 393

Answer:

  A, C

Step-by-step explanation:

The attached graph shows which lines go through the given point. They are ...

  y = 1/2x -1 . . . . 1st selection

  y = -1/6x +3 . . . 3rd selection

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The equations can be found algebraically by substituting the given point in the equation and seeing if the result is a true statement.

  a) 2 = (1/2)(6) -1 = 3 -1 . . . true

  b) 2 = -3(6) . . . . false

  c) 2 = -1/6(6) +3 = -1 +3 . . . true

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2 years ago
Is ketchup a jelllllyyy? Help plsssss
zaharov [31]
No it’s not a jelly or a jam
5 0
2 years ago
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What are the rational numbers of -12,0,35,4.85,√12,√36,19/6,-10/11,1.4949949994
klemol [59]
All of those numbers are rational, except the square root of 12.
6 0
3 years ago
An object is heated to 100°. It is left to cool in a room that
stepladder [879]

Answer:

Step-by-step explanation:

Use Newton's Law of Cooling for this one.  It involves natural logs and being able to solve equations that require natural logs.  The formula is as follows:

T(t)=T_{1}+(T_{0}-T_{1})e^{kt} where

T(t) is the temp at time t

T₁ is the enviornmental temp

T₀ is the initial temp

k is the cooling constant which is different for everything, and

t is the time (here, it's in minutes)

If we are looking first for the temp after 20 minutes, we have to solve for the k value.  That's what we will do first, given the info that we have:

T(t) = 80

T₁ = 30

T₀ = 100

t = 5

k = ?

Filling in to solve for k:

80=30+(100-30)e^{5k} which simplifies to

50=70e^{5k} Divide both sides by 70 to get

\frac{50}{70}=e^{5k} and take the natural log of both sides:

ln(\frac{5}{7})=ln(e^{5k})

Since you're learning logs, I'm assuming that you know that a natural log and Euler's number, e, "undo" each other (just like taking the square root of something squared).  That gives us:

-.3364722366=5k

Divide both sides by 5 to get that

k = -.0672944473

Now that we have a value for k, we can sub that in to solve for T(20):

T(20)=30+(100-30)e^{-.0672944473(20)} which simplifies to

T(20)=30+70e^{-1.345888946}

On your calculator, raise e to that power and multiply that number by 70:

T(20)= 30 + 70(.260308205) and

T(20) = 30 + 18.22157435 so

T(20) = 48.2°

Now we can use that k value to find out when (time) the temp of the object cools to 35°:

T(t) = 35

T₁ = 30

T₀ = 100

k = -.0672944473

t = ?

35=30+100-30)e^{-.0672944473t} which simplifies to

5=70e^{-.0672944473t}

Now divide both sides by 70 and take the natural log of both sides:

ln(\frac{5}{70})=ln(e^{-.0672944473t}) which simplifies to

-2.63905733 = -.0672944473t

Divide to get

t = 39.2 minutes

3 0
3 years ago
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