Algebra 1 grade 9 can anyone me answer this question? the average january temperature in a northern canadian city is 1° fahrenheit. the actual january temperature for that city may be about 5° warmer or colder. write an equation for this scenario. which choice below is correct? [ t - 5 ] = 1 or [ t - 1 ] = 5 what is the minimum temperature for the city above? what is the maximum temperature for the city above?

You are describing where I live only that temperature is a bit low. I'm sorry to say it is neither one.

t_daily = t_average +/- 5

The highest temperature would be

t_daily = t_average + 5

t_daily = 1 + 5

t_daily = 6

The Lowest temperature would be

t_daily = t_average - 5

t_daily = 1 - 5

t_daily = - 4

Sounds about right for December.

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.