well, in short, the students and chaperones are at a 2 : 9 ratio, and we know there are 171 students, how many chaperones then, so long they maintain a 2:9 ratio?

$\bf \cfrac{chaperones}{students}\qquad 2:9\qquad \cfrac{2}{9}\qquad \qquad \cfrac{x}{171}=\cfrac{2}{9}\implies 9x=2(171) \\\\\\ 9x=342\implies x=\cfrac{342}{9}\implies x=38$

5 0

2 years ago

How do you solve 3x-15=7x-10

Fofino [41]

Answer:

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

Exact Form: x=-45

−

Decimal Form:x=-1.25

−

1.25

Mixed Number Form:x=-14

−

5 0

2 years ago

Read 2 more answers

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:

$-.3364722366=5k$