Answer:
Read below.
Step-by-step explanation:
Hey there, grab your solution:
x²+9x+16=0
Sqrt(D)=sqrt(17)
So x is:
(-9+sqrt(17))/2
And (-9-sqrt(17))/2
The short answer is trial and error. The side lengths "3" and "4" can both be substituted for a² and b² but not c² because their value squared is not high enough since 5² is 25. "c²" as to match the longest side because the smaller numbers will cause the equation to not be true. See Below.
a² + b² = c²
3² + 4² = 5²
9 + 16 = 25
25 = 25
______________________________
a² + b² = c²
4² + 3² = 5²
16 + 9 = 25
25 = 25
______________________________
a² + b² = c²
5² + 4² = 3²
25 + 16 = 9
41 ≠ 9
______________________________
a² + b² = c²
3² + 5² = 4²
9 + 25 = 16
34 ≠ 16
Hope this helped!
Answer:
The magnitude of the boat's velocity is 5.66 m/s
Step-by-step explanation:
Given;
Radius of curvature of the circular path, r = 40 m
speed of the boat, v = 0.8 t m/s
The magnitude of the boat's velocity when it has traveled 20 m is ?
a = dv / dt,
v = 0.8 t (differentiate with respect to t)
dv / dt = 0.8 = a
thus, a = 0.8 m/s²
If the boat starts from rest, initial velocity, u = 0
Apply kinematic equation, to solve for the boat velocity at 20 m
v² = u² + 2as
v² = 0 + 2 x 0.8 x 20
v² = 0 + 32
v² = 32
v = √32
v = 5.66 m/s
Therefore, the magnitude of the boat's velocity when it has traveled 20 m is 5.66 m/s
Answer:
Step-by-step explanation:
Given that the time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.
P(completing exam before 1 hour)
= P(less than an hour) = P(X<60)
=P(Z<
)
=0.5-0.34=0.16
i.e. 16% of students completed the standardized exam.
Answer:
(i)Here ∠HGB=∠AGE(vertically opposite angles)
∠HGB=110
∘
Now, for parallel lines l and m, with transversal t.
∠HGB+∠GHD=180
∘
(Angles on the same side of the transversal are supplementary)
110
∘
+∠GHD=180
∘
∠GHD=180
∘
−110
∘
=70
∘
(ii)For parallel lines l and m with transversal a
∠AIE=∠CJI (corresponding angles)
⇒x=100
∘