Answer:
Step-by-step explanation:
Given:
KL ║ NM,
LM = 45
m∠M = 50°
KN ⊥ NM
NL ⊥ LM
To determine: KN and KL
1. Analyzing triangle NLM, we see it is a right triangle due to NL ⊥ LM. In this context,
LM = 45
m∠M = 50°
Consequently,
It is also true that
(angles LNM and M are complementary).
2. Now considering triangle NKL, it also forms a right triangle as KN ⊥ NM. Within this triangle,
(alternate interior angles)
(angles KNL and KLN are complementary).
Thus,
and
<span> The absolute value function exhibits symmetry. Given that the coordinates (–6, –2) and (0, –2) produce the same output, the points are equidistant from the line of symmetry. The value of –3 exists between –6 and 0. Therefore, the x-coordinate of the vertex must be –3, which is the value of </span>h<span>. This indicates that the graph of the parent function shifts 3 units to the left.</span>
The answer must be B Step-by-step explanation:
Answer: The cube root of 10 is 2.1544, using an initial value of -0.003723.
Step-by-step explanation: The Newton-Raphson method is utilized for root finding, and its formula is NR: X=Xo-(f(x)/f'(x)). Before applying this formula, the derivative of the equation must be determined. Given that X =10, this method was implemented to identify the best root to ascertain the cube root of 10 to 5 significant figures. Utilizing software like Excel for quicker iteration calculations is advisable. The found root in this instance was -0.003723.
To determine the values of b that fulfill 3(2b+3)^2 = 36
we start with
3(2b+3)^2 = 36
Divide both sides by 3
(2b+3)^2 = 12
Next, take the square root of both sides
(2b+3)} = (+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3
b1=\frac{\sqrt{12}}{2} -\frac{3}{2}
b1=\sqrt{3} -\frac{3}{2}
b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
Thus,
the solutions for b are
b1=\sqrt{3} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
