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kolbaska11 [484]
3 years ago
13

Describe the sum29-8

Mathematics
2 answers:
almond37 [142]3 years ago
8 0
The sum of the equation is 21
alexandr402 [8]3 years ago
5 0
The answer to the question is the number 17
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Please help!! if you could show work too that would be amazing :))
icang [17]

Answer:

Was there suppose to be a attached image?

5 0
3 years ago
What is the value of this expression? (−4)^4?
Charra [1.4K]
-4^4= -4*-4*-4*-4

-4*-4=16
16*-4=-64
-64*-4= 256

Final answer: 256
5 0
3 years ago
How to solve (8x-30)+(6x+10)=180
lozanna [386]
Add the like terms. So I'll add the variables. 8x+6x is 14x. Now add whole numbers. 30+10 is 40. Our new expression we'll be 14x+40=180. Subtract 40 from 180. 180-40=140. Now we have the expression 14x=140. 140 divided by 14 is 10. So x is 10.
7 0
4 years ago
Im doing decimals greatest to least and i need help with number 12 and 13
erma4kov [3.2K]

(I know there isnt any real problem here to solve, but heres a tip on how to solve greatest to least with decimal problems)

1. Just because a number looks big, doesn't mean it is big

Example: 1.00000000000001 < 1.1

just look at the numbers ^ and dont just hastily read it over assuming that 1.1 is smaller because it "has less digits"

2. Negative numbers are... opposite. and they are less than positive numbers

-3.4 > -4.1

Why is this? Well, if you look on a line, with the point 0 in the middle, you can see that -3.4 is not as far away from 0 as -4.1 is. So the idea is to apply opposite logic for negative numbers

I hope these tips helped!! :D

3 0
3 years ago
Read 2 more answers
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
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