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3 years ago

5

I am between 7,000,000 and 8,000,000. all my digits are odd. all the digits in my thousands period are the same. the sum of my d

igit is 31. what number am i? give as many answers as you can. what strategies did you use to find the mystrey number

2 answers:

lozanna [386]3 years ago

7 0

Answer:

The number can be many numbers like: 7,777,111 or 7,555,333 or 7,111,777

A: 7,777,111

We have four 7's and three 1's

$7*4=28$

$1*3=3$

$28+3=31$

B: 7,555,333

We have one 7, three 5's and three 3's.

$5*3=15$

$3*3=9$

$15+9+7=31$

These also can be 7,333,555 or 7,111,777.

aniked [119]3 years ago

4 0

There are many possibilities so I am going to give 3 answers 777113 777131 777311

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Find the product or type<br>"impossible".<br>3 -5 4 2<br>

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Answer:

The product of $\begin{pmatrix}3&-5\\ \:1&7\end{pmatrix}\begin{pmatrix}4&2\\ \:-4&2\end{pmatrix}$ is $\begin{pmatrix}32&-4\\ -24&16\end{pmatrix}$ .

Step-by-step explanation:

A matrix is a rectangular arrangement of numbers into rows and columns.

Matrix multiplication refers to the product of two matrices.

The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one.

To find the product of $\begin{pmatrix}3&-5\\ \:1&7\end{pmatrix}\begin{pmatrix}4&2\\ \:-4&2\end{pmatrix}$

$\mathrm{Multiply\:the\:rows\:of\:the\:first\:matrix\:by\:the\:columns\:of\:the\:second\:matrix}\\\\\begin{pmatrix}3&-5\end{pmatrix}\begin{pmatrix}4\\ -4\end{pmatrix}=3\cdot \:4+\left(-5\right)\left(-4\right)\\\\\begin{pmatrix}3&-5\end{pmatrix}\begin{pmatrix}2\\ 2\end{pmatrix}=3\cdot \:2+\left(-5\right)\cdot \:2\\\\\begin{pmatrix}1&7\end{pmatrix}\begin{pmatrix}4\\ -4\end{pmatrix}=1\cdot \:4+7\left(-4\right)\\\\\begin{pmatrix}1&7\end{pmatrix}\begin{pmatrix}2\\ 2\end{pmatrix}=1\cdot \:2+7\cdot \:2$

$\begin{pmatrix}3&-5\\ 1&7\end{pmatrix}\begin{pmatrix}4&2\\ -4&2\end{pmatrix}=\begin{pmatrix}3\cdot \:4+\left(-5\right)\left(-4\right)&3\cdot \:2+\left(-5\right)\cdot \:2\\ 1\cdot \:4+7\left(-4\right)&1\cdot \:2+7\cdot \:2\end{pmatrix}$

$\mathrm{Simplify\:each\:element}\\\\\begin{pmatrix}3&-5\\ 1&7\end{pmatrix}\begin{pmatrix}4&2\\ -4&2\end{pmatrix}=\begin{pmatrix}32&-4\\ -24&16\end{pmatrix}$

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4 years ago

Point R is on line segment QS. Given QR = 2 and RS=10, determine the length

Liono4ka [1.6K]

Answer:

Assuming you are talking about segment QS, it's 12 units.

Step-by-step explanation:

10+2=12.

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3 years ago

27 divided by 53 minus 6 plus 53 times -2 squared

mamaluj [8]

Answer:

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\left(-\left(2^2\right)\right)\right)=-\frac{11527}{53}\quad \left(\mathrm{Decimal:\quad }\:-217.49056\dots \right)$

Step-by-step explanation:

Considering the expression

27 divided by 53 minus 6 plus 53 times -2 squared

Which can be written as

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\cdot \left(-\left(2^2\right)\right)\right)$

So, solving the expression

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\cdot \left(-\left(2^2\right)\right)\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$=\frac{27}{53}-6-53\cdot \:2^2$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:6=\frac{6\cdot \:53}{53}$

$=-\frac{6\cdot \:53}{53}+\frac{27}{53}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-6\cdot \:53+27}{53}$

$=\frac{-291}{53}$ ∵ $-6\cdot \:53+27=-291$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-2^2\cdot \:53-\frac{291}{53}$

$=-212-\frac{291}{53}$ ∵ $53\cdot \:2^2=212$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:212=\frac{212\cdot \:53}{53}$

$=-\frac{212\cdot \:53}{53}-\frac{291}{53}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-212\cdot \:53-291}{53}$

$=\frac{-11527}{53}$ ∵ $-212\cdot \:53-291=-11527$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{11527}{53}$

Therefore,

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\left(-\left(2^2\right)\right)\right)=-\frac{11527}{53}\quad \left(\mathrm{Decimal:\quad }\:-217.49056\dots \right)$

Keywords: algebraic expression

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Step-by-step explanation:

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3 years ago

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Answer:

Here's what i know:

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3 years ago

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