Given:
Temperature in the morning = 6°F.
By the late afternoon, the temperature had dropped 9°F.
To find:
The temperature by the late afternoon.
Solution:
Temperature by the late afternoon = Morning temperature - Dropped temperature
Using the given values and the above formula, we get
Temperature by the late afternoon = 6°F - 9°F
Temperature by the late afternoon = -3°F
Therefore, the temperature by the late afternoon is -3°F.
Answer:
216
Step-by-step explanation:
A=2(wl+hl+hw)=2·(12·2+6·2+6·12)=216
As we know that the standard equation of circle is
, where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>
Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>
Now , putting values in our standard equation ;
<em>This is the required equation of </em><em>Circle</em>
Refer to the attachment as well !
Answer:
○
Step-by-step explanation:
This is where your <em>median</em> is near the lower quartile.
I am joyous to assist you anytime.