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svetlana [45]
3 years ago
8

15 of the 25 students Brought their own lunch. What percent of the students brought their own lunch?

Mathematics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

60%

Step-by-step explanation:

1: 15 divided by 25 to get 0.6

2: multiple 0.6 by 100 to get 60%

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In rectangle KLMN, KM = 10x+24 and KN = 64 find x
pishuonlain [190]
Hi there, I don't know if I am right but I will try to solve this, 10x+24=64, subtract 24 from both sides, 10x+24-24=64-24=10x=40, divide both sides by 10, 10x/10=40/10, x=4
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Which of the following is best used to measure the weight of a 18-wheeler truck
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the answer to this would be tons
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What is the value of the expression? −3/4+(1/10÷2/5) Enter your answer, as a fraction in simplest form, in the box.
blagie [28]
-11/20 would be the value of the expression
8 0
3 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

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
Please answer the following question. Also please answer if you actually know it.
Iteru [2.4K]

Given : f(x)= 3|x-2| -5

f(x) is translated 3 units down and 4 units to the left

If any function is translated down then we subtract the units at the end

If any function is translated left then we add the units with x inside the absolute sign

f(x)= 3|x-2| -5

f(x) is translated 3 units down

subtract 3 at the end, so f(x) becomes

f(x)= 3|x-2| -5 -3

f(x) is translated   4 units to the left

Add 4 with x inside the absolute sign, f(x) becomes

f(x)= 3|x-2 + 4| -5 -3

We simplify it and replace f(x) by g(x)

g(x) = 3|x + 2| - 8

a= 3, h = -2 , k = -8



5 0
3 years ago
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