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vladimir1956 [14]
3 years ago
5

Who can do my i-Ready math is only 1 lesson I’m level E pls

Mathematics
1 answer:
DIA [1.3K]3 years ago
5 0
I’ll do it, math or reading?
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What is the solution to 9e-7=7e-11<br> need it now
oee [108]

The answer is e = -2

I hope this helped! If so, please mark brainliest!

Step by step solution :

Step  1  :

Pulling out like terms :

1.1     Pull out like factors :

  2e + 4  =   2 • (e + 2)  

Equation at the end of step  1  :

Step  2  :

Equations which are never true :

2.1      Solve :    2   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

2.2      Solve  :    e+2 = 0  

Subtract  2  from both sides of the equation :  

                     e = -2

8 0
3 years ago
Read 2 more answers
Which properties were used to simplify the following expression? Select all that apply.
Alenkinab [10]

Answer:

A P of Addition

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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
3 years ago
PLS HELP I'LL GIVE U BRAINLIEST!!!!!
umka2103 [35]

Answer:

LOL

Step-by-step explanation:

3 0
3 years ago
Kevin drove 357 miles using 14 gallons of gas. At this rate, how many miles would he drive using 16 gallons of gas?​
Inessa [10]

Answer:

Kevin could drive 408 miles using 16 miles of gas.

Step-by-step explanation:

357 (1/14) = 25.5

25.5 (16) = 408

7 0
3 years ago
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