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BaLLatris [955]
3 years ago
5

Please help me id really appreciate it

Mathematics
1 answer:
loris [4]3 years ago
5 0

Answer:

Step-by-step explanation:

1. 2/5

2. 9/20

3. 0.43

4. 40.2%

5. 42%

6. 0.403

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

I truely hope this is right & i hope this helps you ! <3

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Lucy cuts 4 squares with side length x in. from the corners of a 6 in. by 9 in. cardboard rectangle. She folds the remaining car
Cloud [144]

Answer:

Lucy cuts 4 squares with side length x in. from the corners of a 6 in. by 9 in. cardboard rectangle. She folds the remaining cardboard to make a tray that is x in. high. Write and simplify a polynomial function for the volume V

Step-by-step explanation:

9v

8 0
3 years ago
Hi can someone please help me on my homework
Murrr4er [49]

Answer:

8. yes, ASA

9. yes, SAS

10. yes, AAS

11. no

12. no

13. yes, SSS

Step-by-step explanation:

4 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
4x^2+bx+9=0 has no real number solutions what must be true about b​
timama [110]

b^2-4ac<0 b^2-144<0 (b-12)(b+12)<0

-12<b<12

4 0
3 years ago
Dels anyone know this? (Im just starting algebra now
german

Answer:

1)3

2)77

3)28

4)11

5)7

6)40

7)108

8)7

Step-by-step explanation:

You just have to find what the variable equals.

All these answers are correct

8 0
3 years ago
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