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Setler79 [48]
3 years ago
6

What is 5.27 in Tenths place

Mathematics
2 answers:
kirza4 [7]3 years ago
6 0
It is 5.30 because 27 is rounded to 30.
Aloiza [94]3 years ago
3 0
5.3 would be round to the tenth place.


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hana paid 1,200 for the carpet in her living room. the room has an area of 251.2 square feet. what was her unit cost of carpetin
rosijanka [135]
Itd be $4.78 
1200/251.2=4.77707006, rounded to the nearest cent would be $4.78 :)
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3 years ago
Brainliest to Correct answer! Identify the relationship described below.
kenny6666 [7]

Answer:

Commenalism

Step-by-step explanation:

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3 years ago
Given h(x) = 5x + 2, solve for 2 when h(x) = -8.
san4es73 [151]

Answer:

\huge\boxed{\sf x = -2}

Step-by-step explanation:

\sf h(x) = 5x+2\\\\Put \ h(x) = -8\\\\-8 = 5x+2\\\\Subtract \ 2 \ to \ both \ sides\\\\-8-2 = 5x\\\\-10 = 5x\\\\Divide\ both \ sides \ by \ 5\\\\-10 / 5 = x \\\\x = -2 \\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
8 0
3 years ago
So far Monica has read 56 of a book she has read the same number of pages each day for five days what fraction of the book does
alexandr1967 [171]

Answer:

14/125

Step-by-step explanation:

56 % ÷ 5 days ie 11.2% per day ie 112/1000 or 14/125

4 0
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
3 years ago
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