*the answer has to be a fraction* Thank you if u answer! will give brainliest

PLEASE HURRY Read a text trailer for the movie Harry Potter and the Sorcerer’s Stone. Harry Potter seems at first glance normal,

V125BC [204]

Answer:

Step-by-step explanation:

Which statement in the trailer is the hook?

A) Harry Potter seems at first glance normal, a boy living an unusual life with his dull and rather mean Dursley relatives.

B) However, on Harry’s eleventh birthday, he learns from a mysterious stranger, Rubeus Hagrid, that he is actually a famous wizard.

C) Forbidden by his uncle from seeking out his magical destiny, letters rain down inviting Harry to attend the magic school of Hogwarts.

D) What follows is an adventure story of intrigue, discoveries, and sacrifice.

In order of events:

A, C, B, D

A is where the story begins so it would be the best chose. It should catch you and make you want to know more. Also, the hook is usually the first sentence.

6 0

3 years ago

Using both the rotation matrices earlier in this lesson and your matrix calculator, find each determinant.

Trava [24]

Answer:

both are one

8 0

3 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

A factory robot recently examined some bulbs, of which 2 were flawed and 18 were not. If 30 more bulbs are examined, predict how

Oksanka [162]

Answer:

3 bulbs should be flawed

Step-by-step explanation:

2+ 18 = 20

20 bulbs were examined

2/20 = 1/10 were flawed

looking at the next 30 bulbs

30 * 1/10 = 3