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podryga [215]
3 years ago
8

Elsa went for a drive in her new car. She drove 345.6 miles at a speed of 54 miles per hour. For how many hours did she drive

Mathematics
1 answer:
Kitty [74]3 years ago
6 0

Answer:

6.4

Step-by-step explanation:

You could use a calculator

Or....

Do 345.6/54 and get 6.4

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Based on the graph below, what is the y-intercept?
Sphinxa [80]

Answer:

the y intercept you are looking for is -3

4 0
3 years ago
Si el lunes es el marte del miercoles y el jueves es el viernes del sabado, ¿que dia sera el domingo del lunes?
alex41 [277]

Translated : If Monday is Wednesday's Tuesday and Thursday is Saturday's Friday, what day will Monday be Sunday?

Anwser: Sunday will be monday

5 0
3 years ago
Please answer ASAP!!<br>plssss
34kurt

Answer:

86°

Step-by-step explanation:

b = 29× 2 = 58

d= [180-(86+29)]×2 = 130

a=c=x

a+b+c+d = 360

2x+188= 360

2x= 172

x= 86

a = c = 86°

6 0
3 years ago
Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

5 0
3 years ago
I need help with this
sergij07 [2.7K]

Answer:

What are you trying to find?

Step-by-step explanation:


4 0
4 years ago
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