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3 years ago

Enter the equivalent distance in km in the box.

Mathematics

2 answers:

FinnZ [79.3K]3 years ago

6 0

Answer:

3.5 kilometers

Step-by-step explanation:

35000/100 = 3500

3500/1000=3.5

yan [13]3 years ago

3 0

Answer:

0.350 km

Step-by-step explanation:

Hi there !

35000 cm = 35000/100 m = 350 m

350 m = 350/1000 km = 0.350 km

Good luck !

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Which of these conditions might be true if polygons ABCD and KLMN are similar?

White raven [17]

Answer:

Option A.

Step-by-step explanation:

Properties of the similarity of two polygons,

1). All angles of one polygon will be equal to the angles of the other polygon.

2). Their corresponding sides will be in the same ratio.

If the measures of corresponding angles of KLMN and ABCD are equal, but the lengths of the corresponding sides of ABCD are half those of KLMN,

Then the polygons ABCD and KLMN will be similar.

Option A will be the answer.

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3 years ago

Estimate the value of 196 divided by 0.499

tester [92]

The exact answer would be <span>392.78

An estimated value could be either 390 or 393</span>

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3 years ago

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3 years ago

The average of 4, 9, 16, 25 and x is 2x. What is the value of x? A) 4 B)6 A)9 C)9 D)12

Elodia [21]

The average of 4, 9, 16, 25 and x is 2x means that;
(4+9+16+25+x)/5=2x
Multiply both side by 5
5(54+x)/5=2x(5)
54+x=10x
Subtract x to both side
54+x-x=10x-x
54=9x
Divided 9 to both side
54/9=9x/9
x=6
Check:
Substitute x with 6
(4+9+16+25+x)/5=2x
(4+9+16+25+6)/5=2(6)
60/5=12
12=12; so x=6 which marks B as the correct answer. Have a nice day!

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3 years ago

Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you

adell [148]

Answer:

1) The solution of the system is

$\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) The solution of the system is

$\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

Step-by-step explanation:

1) To solve the system of equations

$\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right$

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

$\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 2: Swap rows 1 and 2

$\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 3: $\left(R_1=\frac{R_1}{6}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 5: $\left(R_2=\frac{R_2}{3}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 6: $\left(R_3=R_3+R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]$

Step 7: $\left(R_3=\left(\frac{6}{37}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 8: $\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 9: $\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 10: Rewrite the system using the row reduced matrix:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) To solve the system of equations

$\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right$

using the row reduction method you must:

Step 1:

$\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 2: $\left(R_1=\frac{R_1}{4}\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 3: $\left(R_2=R_2-\left(2\right)R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 5: $\left(R_2=\left(2\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 6: $\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 7: $\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]$

Step 8: $\left(R_3=\left(- \frac{2}{117}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 9: $\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 10: $\left(R_2=R_2-\left(15\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 11:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

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3 years ago