Answer:
See below
Step-by-step explanation:
Attachment 1 : (a) Remember that it mentions x is the years since 1900. That would mean that the table is a bit different. To create this " new table " simply subtract 1900 from the years provided, and substitute.
To create this equation we will need a regression calculator. The equation will be as follows.
y = 0.125873x - 7.11916 ( note that you can double check this equation be substituting points from the table in the attachment )
(b) 2025 - 1900 = 126 years,
y = 0.125873(125) - 7.11916 = $ 8.614965
Minimum Wage : $ 8.614965
Attachment 2 : The rest of the problems can be solved similarly...
(a) Quadratic Regression Equation : - 0.49311x² + 23.2798x + 996.029
(b) - 0.49311(20)² + 23.2798(20) + 996.029 = 1264.381 mg/cm³
Attachment 3 : (a) Exponential Regression : 9.08292(1.09965)ˣ
(b) 9.08292(1.09965)⁶⁰ = 
( About 2713 recommendations )
Answer:
Sphere Volume = (4/3) * PI radius^3
Sphere Volume = (4/3) * 3 * 3 * 3 * PI =
4 * 3 * 3 * PI =
36 PI
Answer is d 36 PI cubic feet
Step-by-step explanation:
Tienes que buscar el minimo multiplicator comun(?). perdon no se Como se llama xD. pero se juntan cada 15. dias
si la Ultima vez que coinsidieron fue 2 la siguiente Sera 17 de maayo
Answer:
71.63 kilowatts hours
Step-by-step explanation:
We are told in the question:
John's electricity bill costs $20.90 per month plus $1.23 per kilowatt hour.
We are to find, How many kilowatt hours can he use and keep his monthly cost no more than $109.
Step 1
$109 - $20.90
= $88.1
$88.1 is the amount left to spend on killowatts of electricity per hour after removing the normal monthly electricity bill in a month
Step 2
$1.23 = 1kilowatts per hour
$88.1 = y kilowatts per hour
Cross Multiply
= $1.23 × y kilowatts per hour = $88.1 × 1 kilowatts per hour
y kilowatts per hour = $88.1 × 1 kilowatts per hour/ $1.23
= 71.62601626kilowatts hour.
Approximately = 71.63 kilowatts hour
Therefore, John can use 71.63 kilowatts per hour and keep his monthly cost no more than $109
Notice that the condition <em>x</em> ≠ 2<em>πk</em> for (presumably) integer <em>k</em> means cos(<em>x</em>) ≠ ±1, and in particular cos(<em>x</em>) ≠ 1 so that we could divide both sides by (1 - cos(<em>x</em>)) safely. Doing so lets us separate the variables:
(1 - cos(<em>x</em>)) <em>y'</em> - <em>y</em> sin(<em>x</em>) = 0
==> (1 - cos(<em>x</em>)) <em>y'</em> = <em>y</em> sin(<em>x</em>)
==> <em>y'</em>/<em>y</em> = sin(<em>x</em>)/(1 - cos(<em>x</em>))
==> d<em>y</em>/<em>y</em> = sin(<em>x</em>)/(1 - cos(<em>x</em>)) d<em>x</em>
Integrate both sides and solve for <em>y</em>. On the right, substitute <em>u</em> = 1 - cos(<em>x</em>) and d<em>u</em> = sin(<em>x</em>) d<em>x</em>.
∫ d<em>y</em>/<em>y</em> = ∫ sin(<em>x</em>)/(1 - cos(<em>x</em>)) d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>u</em>/<em>u</em>
ln|<em>y</em>| = ln|<em>u</em>| + <em>C</em>
exp(ln|<em>y</em>|) = exp(ln|<em>u</em>| + <em>C </em>)
exp(ln|<em>y</em>|) = exp(ln|<em>u</em>|) exp(<em>C </em>)
<em>y</em> = <em>Cu</em>
<em>y</em> = <em>C</em> (1 - cos(<em>x</em>))
