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vredina [299]
3 years ago
8

2 At Bea's Pet Shop, the number of dogs, d, is initially five less than twice the number of cats, c. If she decides to add three

more of each, the ratio
of cats to dogs will be – Write an equation or system of equations that can be used to find the number of cats and dogs Bea has in her pet shop. Could Bea's Pet Shop initially have 15 cats and 20 dogs? Explain your reasoning. Determine algebraically the number of cats and the number of dogs Bea initially had in her pet shop.​
Mathematics
1 answer:
KatRina [158]3 years ago
6 0

Let D be dogs and C be cats

<em>dogs, d, is initially five less than twice the number of cats, c</em>

D + 5 = 2C

<em>If she decides to add three more of each, the ratio  of cats to dogs will be</em>

D + 8 = 2C + 3

<em>Could Bea's Pet Shop initially have 15 cats and 20 dogs?</em>

Simply plug in the numbers

20 + 5 = 2(15)

This is clearly not true: 25 does not equal 30

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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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