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LekaFEV [45]
3 years ago
13

What equation represents a line that when if graphed, passes through the points (0, -1) and

Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0
Y= 2x-1
Explanation: First, you find slope. If the first point it x-axis point is 0 and the second x-axis point is 1 then you already know the answer is over 1, so you have to figure out how many up you are going on the y-axis from -1 to 1. Which is 2, so your slope is 2/1 or just 2. Next, you find your y-intercept, which is where the line crosses the y-axis, and the easy way to find this is finding what y is equal to when x is equal to 0, in the problem you are told (0, -1) as a point, so they gave you your y-intercept right there, which is -1. Finally you right the equation in slip intercept form (y= mx+b) which is y= 2x-1
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4th option

Step-by-step explanation:

open circle means more than or less than

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What is the highest mulitipul of 2&3
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14 is 12 1/2% of what number?
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2 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
2 years ago
Need ASAP!!!!!!!!!!!!!!!!
Semmy [17]

Answer:

d

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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