Ask question

Ask question

All categories

3 years ago

9

David charges $4 to wash all the windows of a car, inside and out. The amount of money he earns washing the windows must end in

one of what digits? Why? List all the possible digits.

1 answer:

nordsb [41]3 years ago

4 0

Answer:

The possible digits are : 0, 2, 6, 4, 8

Step-by-step explanation:

Money charged by David to wash the windows of a car = $4

Let total number of cars he washed be x

Now, Total money earned by washing windows of a car = Money charged for washing all windows of one car × Total number of cars washed

⇒ Total money earned = 4 × x

So, the amount will always end in the digits which comes at the end of multiples of 4 because the amount will be always in the multiples of 4

⇒ 4 × 1 = 4 , 4 × 2 = 8 , 4 × 3 = 12 , 4 × 4 = 16 , 4 × 5 = 20 ......

So, the possible digits are : 0, 2, 6, 4, 8

You might be interested in

Find the LCM of n^3 t^2 and nt^4.<br><br> A) n 4t^6<br> B) n 3t^6<br> C) n 3t^4<br> D) nt^2

patriot [66]

Answer:

The correct option is C.

Step-by-step explanation:

The least common multiple (LCM) of any two numbers is the smallest number that they both divide evenly into.

The given terms are $n^3t^2$ and $nt^4$ .

The factored form of each term is

$n^3t^2=n\times n\times n\times t\times t$

$nt^4=n\times t\times t\times t\times t$

To find the LCM of given numbers, multiply all factors of both terms and common factors of both terms are multiplied once.

$LCM(n^3t^2,nt^4)=n\times n\times n\times t\times t\times t\times t$

$LCM(n^3t^2,nt^4)=n^3t^4$

The LCM of given terms is $n^3t^4$ . Therefore the correct option is C.

8 0

3 years ago

Read 2 more answers

HELP FAST ITS TO RPROVE MY TEACHER WRONG

OlgaM077 [116]

Answer

use goo gle

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

John has a rope that is 10 yards long. HoW long is his rope in inches?

jek_recluse [69]

Answer:

360 in.

Step-by-step explanation:

1 yd = 36 in.

10 yd = 10 * 1 yd = 10 * 36 in. = 360 in.

7 0

2 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago

If someone can answer this, I will give you 90 points. How fast in terms of miles does the speed of light travel at?

tangare [24]

Answer:

670,616,629mph

Step-by-step explanation:

I think this is the answer

7 0

3 years ago

Read 2 more answers