<h3>3 answers: Choice A), Choice C), Choice D)</h3>
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Explanation:
We have a triangle with sides a,b,c where a = 3 and b = 3 are given. The side c is unknown, but using the variation of the triangle inequality theorem allows us to say the following:
b-a < c < b+a
3-3 < c < 3+3
0 < c < 6
The missing side c is between 0 and 6 units long, but cannot be equal to 0 and can't be equal to 6 either. Therefore, the answers must be A) 4, C) 5, and D) 2. All of these values satisfy the inequality 0 < c < 6.
Something like c = 12 is too large. I recommend cutting out two slips of paper that are 3 inches long, and you'll find it impossible to form a triangle that has a third side of 12 inches. The longest side you can form is 3+3 = 6 inches, but you wouldn't have a triangle at that point (it would be a straight line instead). This is a visual way to rule out choice E, and it also rules out choices B and F for similar reasoning.
Answer: A,C,D
Step-by-step explanation:
Answer:
y= 1.6
You make 2 a decimal by adding .0 at the end (2.0). Then you subtract 0.4 by 2.0 which equals 1.6.
Answer:
NF = 25
Step-by-step explanation:
Since ∆NKF ~ ∆LZF, the ratio of their corresponding side lengths would also be the same.
This means that:
KF/ZF = NF/LF
KF = x + 3
ZF = 4
NF = 15 + x + 3 = x + 18
LF = x + 3
Plug in the values into the equation
(x + 3)/4 = (x + 18)/(x + 3)
Cross multiply
(x + 3)(x + 3) = (x + 18)(4)
x² + 3x + 3x + 9 = 4x + 72
x² + 6x + 9 = 4x + 72
x² + 6x + 9 - 4x - 72 = 0
x² + 2x - 63 = 0
Factorize to find x
x² + 9x - 7x - 63 = 0
x(x + 9) -7(x + 9) = 0
(x + 9)(x - 7) = 0
x + 9 = 0 or x - 7 = 0
x = -9 or x = 7
We'd use the positive value of x, which is 7.
Therefore, x = 7.
✅NF = 15 + (x + 3)
Plug in the value of x
NF = 15 + (7 + 3) = 15 + 10
NF = 25
Answer:
0.128rad/sec
Step-by-step explanation:
Let x represent the between the man and the point on the path
θ = the angle
dx/dt = 4 ft/s
dθ/dt = rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight
tan θ = x/20 ft
Cross Multiply
20tan θ = x
dx/dt = 20sec² θ dθ/dt
dθ/dt = 1/20 × cos² θ dx/dt
dθ/dt= 1/20 × cos² θ × 4
dθ/dt = 1/5 × cos² θ
Note : cos θ = 4/5
dθ/dt = 1/5 × (4/5)²
dθ/dt = 16/125
dθ/dt = 0.128rad/sec
