Answer:
60 °F
Step-by-step explanation:
The function can be put into vertex form:
T(x) = 0.264(x² -18x) +81 . . . . factor the leading coefficient from 1st 2 terms
Now, we add the square of half the x-coefficient inside parentheses and subtract the same quantity outside parentheses.
T(x) = 0.264(x² -18x +81) +81 -0.264·81
T(x) = 0.264(x -9)² +59.616
The low temperature for the day was approximately 59.6 °F, which rounds to 60 °F.
The answer is -2 when you simplify
Not enough information to properly answer this... hope this helped though
Since he descended 12 meters, we subtract this from the overall height of Mount Ka'ala, so then we are only calculating how high ABOVE the sea level it is.
1232 - 12 = 1220
The height of Mount Ka'ala is therefore 1,220 meters.
To calculate how much a fifth of Mount Ka'ala is (since the ranger station is 2/5's up), we would divide this number by 5
1220 ÷ 5 = 244
Since ONE fifth of the height is 244 meters, TWO fifths would be double that amount.
244 x 2 = 488
488 meters.
The ranger station is 488m above sea level.
Answer:
<h2>
∠PQT = 72°</h2>
Step-by-step explanation:
According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.
Also from the diagram, ∠OQP + ∠PQT = ∠OQT;
∠PQT = ∠OQT - ∠OQP
Given ∠OQP = 18° and ∠OQT = 90°
∠PQT = 90°-18°
∠PQT = 72°