Ask question

Ask question

All categories

4 years ago

5

Tom has a collection of 30 CDs and Nita has a collection of 18 CDs. Tom is adding 1 CD a month to his collection while Nita is a

dding 5 CDs a month to her collection. Find the number of months after which they will have the same number of CDs. show your work

2 answers:

timurjin [86]4 years ago

7 0

Answer:

3 months.

Step-by-step explanation:

Let x represent the number of months.

We have been given that Tom has a collection of 30 CDs and he is adding 1 CD a month to his collection.

So number of CDs in Tom's collection after x months would be: $30+x$ .

We are also told that Nita has a collection of 18 CDs. Nita is adding 5 CDs a month to her collection. So number of CDs in Nita's collection after x months would be: $18+5x$ .

To solve for number of months, when both will have same number of CDs, we will equate both expressions as:

$18+5x=30+x$

$18+5x-x=30+x-x$

$18+4x=30$

$18-18+4x=30-18$

$4x=12$

$\frac{4x}{4}=\frac{12}{4}$

$x=3$

Therefore, after 3 months they will have the same number of CDs.

GrogVix [38]4 years ago

6 0

30:18,
31:23
32:28
33:33
It will be three months before they have the same amount of CDs

You might be interested in

What is the length of TH, when T (10,7), and H (-4,3)?

drek231 [11]

Answer:

TH=14.56

Step-by-step explanation:

Given: T(10,7) and H(-4,3)

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\text{Length of TH }=\sqrt{(-4-10)^2+(3-7)^2}$

$=\sqrt{14^2+4^2}$

$=\sqrt{196+16}$

$=\sqrt{212}$

$\approx 14.56$

Hence, The length of TH $\sqrt{212}\approx 14.56$

5 0

3 years ago

Suppose that IQ scores have a bell-shaped distribution with a mean of 100 and a standard deviation of 14. Using the empirical ru

Brut [27]

Answer:

99.7%

Step-by-step explanation:

Bell shaped curve by empirical rule is that if the standard deviation from the mean (left to right) is 1 deviation then about 68% of the data is in that range if it’s 2 standard deviations away then it’s 95% and 3 deviations away is 99.7%

142-100=42=3(14)

100-58=42=3(14)

Therefore the data is three standard deviations from the mean and this there is 99.7% of the data in this range.

3 0

3 years ago

In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with

Zarrin [17]

First check the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -2, so the characteristic solution is

$y_c = C_1e^{-2t} + C_2te^{-2t}$

For the particular solution corresponding to $12te^{-2t}$ , we might first try the <em>ansatz</em>

$y_p = (At+B)e^{-2t}$

but $e^{-2t}$ and $te^{-2t}$ are already accounted for in the characteristic solution. So we instead use

$y_p = (At^3+Bt^2)e^{-2t}$

which has derivatives

${y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}$

${y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}$

Substituting these into the left side of the ODE gives

$(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}$

so that 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 0.

For the second solution corresponding to $-8t-12$ , we use

$y_p = Ct + D$

with derivative

${y_p}' = C$

${y_p}'' = 0$

Substituting these gives

$4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12$

so that 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -1.

Then the general solution to the ODE is

$y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1$

Given the initial conditions <em>y</em> (0) = -2 and <em>y'</em> (0) = 1, we have

$-2 = C_1 - 1 \implies C_1 = -1$

$1 = -2C_1 + C_2 - 2 \implies C_2 = 1$

and so the particular solution satisfying these conditions is

$y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1$

or

$\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}$

7 0

3 years ago

Somebody can help me please

MissTica

It’s 25. Plug -2 into g, which is -5. Then plug -5 into f, which is 25.

6 0

3 years ago

3.2% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely ac

Ivenika [448]

Really don’t kno it’s a hard question

6 0

3 years ago