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natima [27]
3 years ago
8

Solve thus system by substitution​

Mathematics
2 answers:
damaskus [11]3 years ago
8 0

Answer:

(-4,6)

Step-by-step explanation:

y=6

5x+5(6)=10

5x+30=10

5x=-20

x=-4

(x,y)

(-4,6)

notka56 [123]3 years ago
6 0
You just plug in 6 to where y is in the equation because it tells you what number y is. and then u just solve for x

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PLZ HLP Given the equation 5x + y = 7, which equation below would cause a consistent-independent system?
DanielleElmas [232]

Answer:

the equation D ) would cause a consistent-independent system.

Step-by-step explanation:

A ) 5 x + y = 7  /*( -2 )

    10 x + 2 y = 14

    --------------------

    - 10 x - 2 y = - 14

      10 x + 2 y = 14

   ------------------------

       0 x = 0  ( Dependent system )

B ) 5 x + y = 7  / *  3

    - 15 x - 3 y = - 6

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    15 x + 3 y  = 21

    - 15 x - 3 y = - 6

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        0 x = 15 ( Inconsistent system )

C )   5 x + y = 7

       5 x + y = - 7  / * ( - 1 )

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       5 x + y = 7

      - 5 x - y = 7

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D ) 5 x + y = 7  / * ( - 2 )

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7 0
3 years ago
Determine the following<br>quadratic equations have real<br>roots<br><br>1) 3x² - √2x-√3​
IceJOKER [234]

Step-by-step explanation:

√3 x² - 2x - √3 = 0

√3 x² - 3x + x - √3 = 0

√3 x(x - √3) + 1(x - √3) = 0

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3 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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