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4 years ago

What 3-digit number when doubled is, by value, one more than the same 3-digit number reversed? There can be two answers.

Mathematics

1 answer:

Lubov Fominskaja [6]4 years ago

3 0

Let ones digit = z

Let tens digit = y

Let hundreds digit = x

Then three digit number xyz will be equivalent to (100x+10y+z)

If we reverse the digits then new number zyx will be equivalent to (100z+10y+x)

2(100x+10y+z)=(100z+10y+x)+1

200x+20y+2z=100z+10y+x+1

199x+10y-98z-1=0

There are no more conditions given so that we can use them to minimize calculation in order to find the answer.

So now we have only choice to plug random numbers for x,y,z which can vary from 0 to 9.

Working as explained above gives x=3, y=9, z=7 is satisfying the above equation.

Hence required number is 397.

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How do you find the fraction of 0.35

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Answer:

$\frac{7}{20}$

Step-by-step explanation:

0.35 is the same as 35/100

which can be simplified to 7/20

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What is the solution to this equation? √x=7 A. x=√ 49 B. x=√7 C.x=14 D. x= 49

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Let's not look at the option and try to solve it.
√x=7
Let's look at this. Square roots are the number that is multiplied by itself to make that number. For example, 2 is the square root of 4, 3 is the square root of 9.
We need to multiply 7 by itself to find x.
7×7=49
So, D. =49.

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3-5y= 2x

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Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$