Answer:
C
Step-by-step explanation:
Therefore, the correct answer is choice C. Hope this helps!
Answer:
Step-by-step explanation:Since we want to know how much time elapsed since the object was launched until it ht the ground. At the ground, the height is 0.
Therefore the equation h(t) = 80t – 16t2 becomes 0= 80t – 16t2
Solving the equation for t gives 80t =16t2 or 16t=80 or t=5sec
The object hits the ground 5 seconds after it is launched.
7(thousands)
8(hundreds)
5(tens)
4(ones)
.(decimal point)
2(tenths)
0(hundredths)
9(thousandths)
Hope this helps! I would appreciate it if I got brainliest! Thanks! =D
Step-by-step explanation:
(4s + 2) × (5s² + 10s +3)
we know immediately the first and the last term of the result, because there is only one operation to build them :
20s³ and 6.
so, the first answer option is already eliminated.
let's multiply
4s×5s² + 4s×10s + 4×3s + 10s² + 20s + 6
20s³ + 40s² + 12s + 10s² + 20s + 6
20s³ + 50s² + 32s + 6
so, the last answer option is correct.
<em>4℉.</em>
What we know about Degrees is that there is a<em> </em><u><em>Positive</em></u> type and a <u><em>negative</em></u> type.
(i.e: 30℉ is <u><em>positive</em></u> and -30℉ is <u><em>negative</em></u>.)
If the temperature was -4℉ at 7AM, then it is negative. If it goes up by an amount that is more than 4 then that negative will go up to a positive temperature. In this case: At 9AM it was 8° <u><em>warmer</em></u>.
<u><em>Warmer</em></u><em> is a </em><u><em>keyword</em></u><u>.</u> If it is warmer by an amount, Negative temperatures <u><em>will go up to a positive</em></u> and positive temperature <u><em>will just go up</em></u>. If it gets cooler, negative temperatures <u><em>will go down further</em></u> and positive temperatures <u><em>will go down to a negative</em></u>.
So lets work out this problem with our newfound knowledge.
-4° F at 7AM
8° warmer at 9AM
-4 + 8 = 4.
<em>The temperature was 4° at 9AM.</em>
-Snooky
