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3 years ago

PLEASE HELP ME!! I NEED TO GET THIS DONE

Mathematics

1 answer:

Rainbow [258]3 years ago

5 0

Answer:

Step-by-step explanation:

Dilation. Since it's only the size of the figure that's changing

Height of the smaller one: 4

Width of the smaller one: 6

Height of the bigger one: 6

Width of the bigger one: 9

6/4 = 1.5

9/6 = 1.5

Since both 6/4 and 9/6 was 1.5, that means that the Width and Height was proportional and that the two trapezoids is at a ratio o:

1 : 1.5

One unit of the smaller one is 1.5 units on the bigger one.

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Help i will mark you brainliest!

Morgarella [4.7K]

Since x+y equals 5 that means x is 3 and y is 2 because 3+2=5. So now plug it in, 2(3)+3(2). Two multiplied by three is 6 and three times two is 6. So now it’s 6+6 which equals 12. So it’s 2(3)+3(2)=12

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3 years ago

Altitude above sea level is given in positive values and below sea level is given in negative values. Which situation describes

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3 years ago

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
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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

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