4a5+28+6b7-30
4a5+6b7-2<—- this is a to the power of 5 , b to the power of 7
(Is that 5a or a to the power of 5 )
20a+28+42b-30
20a+42b-2<—— this is 5a , 7b
Answer:
Port r is 100° from Port p and 26km from Port p
Step-by-step explanation:
Lets note the dimension.
From p to q = 15 km = a
From q to r = 20 km= b
Angle at q = 50° + 45°
Angle at q = 95°
Ley the unknown distance be x
Distance from p to r is the unknown.
The formula to be applied is
X²= a²+ b² - 2abcosx
X²= 15² + 20² - 2(15)(20)cos95
X²= 225+400-(-52.29)
X²= 677.29
X= 26.02
X is approximately 26 km
To know it's direction from p
20/sin p = 26/sin 95
Sin p= 20/26 * sin 95
Sin p = 0.7663
P= 50°
So port r is (50+50)° from Port p
And 26 km far from p
F(x) = 9x²<span> - 5x + 2
</span>f(-2) = 9(-2)² - 5(-2) + 2
<span>f(-2) = 36 + 10 + 2
</span><span>f(-2) = 48</span>
150 to 60. But the more commen why to do that were to simplify it.
