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3 years ago

Help please help me out with this (old math)

Mathematics

2 answers:

shutvik [7]3 years ago

6 0

6 7/12 I hope this helps ya out

Maurinko [17]3 years ago

5 0

Answer: $6\frac{7}{12}$

Step-by-step explanation:

$\frac{34}{12} + \frac{45}{12} = \frac{79}{12}$

$\frac{79}{12} = 6\frac{7}{12}$

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3 years ago

Write the equation of the line that passes through the given points in slope intercept form

shepuryov [24]

Answer:

y=3x

Step-by-step explanation:

To put this in slope intercept form, let’s first find the slope

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5 0

2 years ago

Read 2 more answers

PROBLEM 2<br> What is the greatest common factor of 15 and 30?

seropon [69]

Answer: 15

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

zloy xaker [14]

Answer:

$t=-20ln\left(\dfrac{1}{3}\right)$

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

$A=45e^{-0.05t}$

We want to determine the value of t for which the area, A(t)=15 square kilometers.

$15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)$

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.

7 0

3 years ago