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exis [7]
3 years ago
11

Rosie did the following problem. (7-1) 2 ÷ 6 + 4 · 3 6 2 ÷ 6 + 4 · 3 36 ÷ 6 + 4 · 3 6 + 4 · 3 10 · 3 30 Which statement best des

cribes her answer?
Rosie is incorrect because she needs to do the exponent before subtracting. Rosie is incorrect because she needs to multiply before she divides. Rosie is incorrect because she needs to multiply before she adds. Rosie is correct.
Mathematics
1 answer:
Travka [436]3 years ago
4 0
Answer: Rosie is incorrect because she needs to do the exponent before subtracting.
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832 x156 show your work​
Sedaia [141]

Answer:

129792

Step-by-step explanation:

         156

     ×  832

    ----------

         312

      468

+  1248

---------------

  129792

6 0
2 years ago
The temperature is 3C at sunrise. By noon ,the temperature changes by 5C. What could be the temperature at noon?
kaheart [24]

Answer:

it could either be 8C or -2C

Step-by-step explanation:

7 0
3 years ago
Please help! find the value of x
Serga [27]
I needed points sorry
4 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
I need to put equations into slope intercept and i need big help its a test grade
creativ13 [48]

Answer:

4.

y= 2/3x + 3

Step-by-step explanation:

By using the distributive property on the right side of the equation you get 2/3x + 2. Then you add 1 to the right side to cancel out the (-1) on the left.

5 0
3 years ago
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