Athermometer read -17° f at 1: 00 am. by 2: 30 am the temperature has dropped 9.1 degree f. what was the temperature at 2: 30 am?

The temperature was -26.1° f at 2:30 am.

Step-by-step explanation:

* In this type of problems, we must think about

- What is the meaning of wards;

 above and below or

 raised and dropped or

 increased and decreased

- For above , raised , increase

 we will add the value of them to the initial value

- For below , dropped , decreased

 we will subtract the value of them from the initial value

* In our problem:

- The reading of thermometer is -17° f at 1:00 am.

- The temperature has dropped 9.1° f when the time was 2:30 am.

* Lets use the explanation above

- We will subtract 9.1 from -17

∴ -17 - 9.1 = -26.1° f

* The temperature was -26.1° f at 2:30 am.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

the answer is ! ?