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kolezko [41]
3 years ago
14

for the decimal 0.555 the value of the digit in the tens place is how many times as much as the value of the digit in the thousa

nds place
Mathematics
2 answers:
NNADVOKAT [17]3 years ago
7 0
The value of the digit in the tens place is 100 times as much as the value of the digit in the thousands place.
postnew [5]3 years ago
3 0

Answer:

10 times.

Step-by-step explanation:

Let the value of digit in tenths place be x times of value of digit in thousandths place.

We are asked to find the value of the digit in the tenths place is how many times as much as the value of the digit in the thousandths place.

0.555

Tenths: 5

Thousandths: 5

\text{Five tenths}=0.5=\frac{5}{10}

\text{Five thousandths}=0.05=\frac{5}{100}

\frac{5}{100}\cdot x=\frac{5}{10}

\frac{100}{5}*\frac{5}{100}\cdot x=\frac{5}{10}*\frac{100}{5}

x=10

Therefore, the value of the digit in the tenths place is 10 times as much as the value of the digit in the thousandths place.

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Price of a mobile phone including 10% GST is rs.16,600. Find the p[rice of mobile phone before gst was added
Nimfa-mama [501]

Answer:

16,600 = 110%, you can do this in your head minus 1,660 GST (whatever that is) 14,940

Step-by-step explanation:

6 0
3 years ago
Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?
Ede4ka [16]

Answer:

The solution is x=2,\ y=0,\ z=-1.

Step-by-step explanation:

You are given the system of three equations:

\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.

So,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.

Multiply the third equation by 23 and subtract it from the second equation:

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.

Hence,

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1

Substitute it into the second equation:

23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0

Substitute them into the first equation:

3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2

The solution is x=2,\ y=0,\ z=-1.

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