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4 years ago

How do the 3's compare in 350 and 403?

Mathematics

1 answer:

Gnesinka [82]4 years ago

7 0

In 350, the 3 is in the hundred's place

in 403, the 3 is in the one's place

hope this helps

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At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

4 years ago

Solve the inequality for u<br>-5u+27 [less than or equal to] 2

baherus [9]

Answer:

5≤u

Step-by-step explanation:

Put the words and numbers into an equation:

-5u+27≤2

Put all like terms on one side:

27-2≤5u ---> Do you see what I did there? Now you don't have to work with negatives. :)

Simplify:

25≤5u

Solve:

5≤u

6 0

3 years ago

Solving multi-step equations 1) p-1=5p+3p-8

Feliz [49]

P=1. I hope this helps. :)

6 0

3 years ago

Read 2 more answers

Bella works at the grocery store and has 453 oranges to fit into 17 boxes. She fills all of the boxes with exactly the same numb

LUCKY_DIMON [66]

Answer:

It would be 7701

Step-by-step explanation: hope this helps!

6 0

3 years ago

Read 2 more answers

Claire traveled 701 miles. She drove 80 miles every day. On the last day of her trip she only drove 61 miles. Write and solve an

mina [271]

Answer:

Claire traveled for 9 days.

Step-by-step explanation:

Given:

Total Distance traveled = 701 miles

Distance traveled each day = 80 miles

Distance traveled on last day = 61 miles

We need to find the number of days Claire traveled.

Solution:

Let the number of days Claire traveled be denoted by 'd'.

Now we can say that;

Total Distance traveled is equal to sum of Distance traveled each day multiplied by number of days and Distance traveled on last day.

framing in equation form we get;

$80d+61=701$

Now Subtracting both side by 61 using Subtraction Property of Equality we get;

$80d+61-61=701-61\\\\80d = 640$

Now Dividing both side by 80 we get;

$\frac{80d}{80}=\frac{640}{80}\\\\d=8$

Hence Claire traveled 80 miles in 8 days and 61 miles on last day making of total <u>9 days</u> of travel.

7 0

4 years ago