Hi
f(x) = g(x) if -x²+3x-2 - ( -x+1) = 0
-x² +3x-2 +x-1 = 0
-x² +4x -3 = 0
To solve, tou have to use the general method of resolution of a quadratic fonction.
To determine if it's has a solution in R, let's calculate Δ
Δ = (4)² - 4 * (1) *(-3)
Δ = 16 +12
Δ= 28
as Δ≥ 0 so the function allow two solution within R
so S 1 = ( -4 +√28) / 2 S 2 = (-4 -√28 ) /2
S1 = ( -4 + 2√7) /2 S2 = (-4 - 2√7) /2
S1 = (2 (-2 +√7) /2 S2 2 (-2 -√7) /2
S1 = -2 +√7 S2 = -2 -√7
So the two function are equal twice. one for x = -2 +√7 and second x = -2-√7
Step-by-step explanation:
In order to be like terms, the variables and exponents must be the same.
3a can be combined with 14a and 4a.
4b can be combined with 3b and 16b.
a² cannot be combined with any of the terms.
More than two is your answer
The way to answer this question is to first make 52% into a decimal:
52% = 0.52
so now we will simply multiply:
84
x.52
once you have multiplied these numbers it comes out to be 43.68.
So your answer is 43.68.
Hope this answer helps! feel free to ask any additional questions :)
Answer:
<h2>b = 15°</h2>
Step-by-step explanation:
If Pq = RQ then ΔPQR is the isosceles triangle. The angles QPR and PRQ have the same measures.
We know: The sum of the measures of the angeles in the triangle is equal 180°. Therefore we have the equation:
m∠QPR + m∠PRQ + m∠RQP = 180°
We have
m∠QPR = m∠PRQ and m∠RQP = 60°
Therefore
2(m∠QPR) + 60° = 180° <em>subtract 60° from both sides</em>
2(m∠QPR) = 120° <em>divide both sides by 2</em>
m∠QPR = 60° and m∠PRQ = 60°
Therefore ΔPRQ is equaliteral.
ΔPSR is isosceles. Therefore ∠SPR and ∠PRS are congruent. Therefore
m∠SPR = m∠PRS
In ΔAPS we have:
m∠SPR + m∠PRS + m∠RSP = 180°
2(m∠SPR) + 90° = 180° <em>subtract 90° from both sides</em>
2(m∠SPR) = 90° <em>divide both sides by 2</em>
m∠SPR = 45° and m∠PRS = 45°
m∠PRQ = m∠PRS + b
Susbtitute:
60° = 45° + b <em>subtract 45° from both sides</em>
15° = b
