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2 years ago

Which formula expresses the number of months, m, that there are in y years in terms of y?

2 answers:

7 0

The correct answer is: [C]: "12y" .
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"12y" = 12 multiplied by the number of years; since each year has "12 months".
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Andru [333]2 years ago

5 0

The answer is c) 12y

When trying to find the number of months in y amount of years, you multiply y by 12, which would be represented by the equation 12y

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the diameter of a football is five times the diameter of a cricket ball ratio of surface area of football and cricket ball is

LenKa [72]

Answer:

25 : 1

Step-by-step explanation:

The diameter of the football is 5 times the diameter of the cricket ball. This means that the radius of the football is also 5 times the radius of the cricket ball.

Let the radius of the football be R.

Let the radius of the cricket ball be r.

This implies that:

R = 5r

Both balls are spherical.

The surface area of the cricket ball is given as:

$a = 4\pi r^2$

The surface area of the football is given as:

$A = 4\pi R^2$

$=>A = 4\pi (5r)^2\\\\A = 100\pi r^2$

Therefore, the ratio of their surface areas is:

A : a

$100\pi r^2 : 4\pi r^2$

= 100 : 4

25 : 1

5 0

2 years ago

The general form of the equation of a circle is Ax2 + By2 + Cx + Dy + E = 0, where A = B 0. If the circle has a radius of 3 unit

Lynna [10]

The correct answer is b.

8 0

2 years ago

Read 2 more answers

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$