Domain 0≤x≤12 this means the domain (x values) are between x=0 and x=12 and you started measuring the degree of temperature starting at a time of 0 and stopped measuring the temperature at a time of 12 hours
Range -4≤y≤4 this means the range (y values) are between y=-4 and y=4. The range is referring to the degree in celcius based on what time you measured it
Hope this helps!
Answer:
no four is not better than 7679_727=2873
Step-by-step explanation: 89 is 48-23
Answer:
no idea sorry!!
Step-by-step explanation:
i think its the third but not sure
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
Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.
so.. that'd be the V(x) for such box, now, where is the maximum point at?
now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.
so, the volume will then be at
