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joja [24]
2 years ago
9

If the 6am temperature in Durango, Colorado was -8° F and the temperature rose 15° by 2pm, then what would the temperature be at

2pm?
Mathematics
1 answer:
blsea [12.9K]2 years ago
4 0
Hello there! Given the temperature at 6am in Durango, Colorado was -8° Fahrenheit:

-8° + 15° = temperature at 2pm

• When adding a positive to a negative, we are subtracting the numbers and taking the sign of the greater number. The equation is represented as so:

(-) + (+) = (+)

The general rule corresponding to this equation is that the answer acquires the sign of the greater number. In this case, because 15 is greater than 8, our answer will be a positive number.

-8 + 15 = 7

From 6am to 2pm, the temperature would have risen from -8° Fahrenheit to 7° Fahrenheit. If you need any extra help, let me know and I will gladly assist you.
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The school store sells pencils and erasers.
garri49 [273]

Answer:

$0.18

Step-by-step explanation:

we need to set up two simultaneous equations

using variables, pencils = p and erasers = e

3 pencils and 2 erasers, the cost is $0.76

3p + 2e = $0.76

2 pencils and 4 erasers, the cost is $1.04.

2p + 4e = $1.04

we now have

3p + 2e = $0.76

2p + 4e = $1.04

to make the number of erasers the same, multiply the first equation by 2 to give 4e

2(3p + 2e = $0.76)

6p + 4e = $1.52

now we have the same number of erasers for both equations

6p + 4e = $1.52

2p + 4e = $1.04

subtract across: 6p - 2p = 4p, 4e - 4e = 0, $1.52 - $1.04 = $0.48

we are left with 4p = $0.48

divide both sides by 4

p = $0.12

1 pencil = $0.12

go back to the start of both equations and use one of them to find 1 eraser. I'll use 3p + 2e = $0.76

input $0.12 in p

3($0.12) + 2e = $0.76

$0.36 + 2e = $0.76

subtract $0.36 on both sides

2e = $0.76 - $0.36

2e = $0.40

divide 2 on both sides

e = $0.20

1 eraser = $0.20

How much more does 1 eraser cost than 1 pencil?

we now know 1 pencil = $0.12 and 1 eraser = $0.20

find the difference between them

$0.20 - $0.12 = $0.18

final answer= $0.18

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3 years ago
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otez555 [7]
I think that the answer is 390
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Do logarithmic functions have horizontal asymptotes?
Marta_Voda [28]

9514 1404 393

Answer:

  not generally

Step-by-step explanation:

The basic log function has a vertical asymptote where its argument is zero. It has a positive slope for all positive arguments. It tends to infinity as the argument tends to infinity. There is no horizontal asymptote.

__

A log function can be made part of a larger expression in a way that would give the expression a horizontal asymptote. For example, the function ...

  f(x) = 1/log(x)

has a horizontal asymptote at f(x) = 0. (This asymptote is approached very slowly.)

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Donut $2.50<br> Cupcake $4.25<br> Ice cream $3.80<br><br> I need help finding the equation
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