No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.
Consider polynomial equations x^2 = 3 and x^2 = 1.
Obviously they have real solutions.
Subtract the two polynomial equations:
(x^2 - x^2) = (3 - 1)
0 = 2...
We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.
Obviously 0 = 2 has no real solution.
draw a rectangle triangle :
sin B=8/17 that means that de biggest side are 17 ("hipotenusa"), and another side is 8. Applying de theorem of pythagoras:
The length of AC is 16 km.
Solution:
Given data:
AB = c = 14 km, ∠A = 30° and ∠B = 89°
AC = b = ?
<u>Let us first find angle C:</u>
<em>Sum of all angles in a triangle = 180°</em>
m∠A+ m∠B + m∠C = 180°
30° + 89° + m∠C = 180°
119° + m∠C = 180°
Subtract 119° from both sides, we get
m∠C = 61°
<u>To find the length of AC:</u>
<em>Using sine formula:</em>
Substitute the given values in the formula.
Multiply by sin 89° on both sides.
The length of AC is 16 km.
From all the steps below, we have been able to prove that; 1 - cot23° = 2/(1 - cot22°).
<h3>How to prove trigonometric functions?</h3>
We want to prove that 1 - cot23° = 2/(1 - cot22°).
We will prove it using the trigonometric expression
cot(22° + 23°) = cot45°
Using trigonometric identities, we can rewrite as;
(cot22° cot23° - 1)/(cot22° + cot23°) = 1
Cross multiply to get;
cot22° cot23° - 1 = cot22° + cot23°
Rearrange to get;
cot22° cot23° - 1 - cot22° - cot23° =0
⇒ cot22° cot23° - 1 - cot22° - cot23° + 2 =2
⇒ cot22° cot23° + 1 - cot22° - cot23° =2
⇒ cot22° cot23° - cot22° - cot23° + 1 = 2
⇒ cot22° (cot23° - 1) - 1 (cot23° - 1) = 2
⇒ (cot22° - 1) (cot23° - 1) = 2
Divide both sides by (cot23° - 1) to get;
cot23° - 1 = 2/(cot22° - 1)
⇒ 1 - cot23° = 2/(1 - cot22°).
Read more about trigonometric proof at; brainly.com/question/7331447
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Answer:
the answer is option B
Step-by-step explanation:
answer is option B