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3 years ago

Select the correct answer.

Mathematics

2 answers:

erica [24]3 years ago

8 0

Answer:

there is 1/2 a chance you will roll an odd number.

Stella [2.4K]3 years ago

6 0

Answer:

I think the correct answer is C. 1/2

Step-by-step explanation:

Why I think this is because it makes the most sense to me.

I hope this helps! :)

Please give brainliest if this is correct!

P. S.

I am so very very sorry if this is wrong.

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This is due tomorrow if someone could help me that would be great <3

kifflom [539]

Hope this helped!

First Row

$\frac{5}{18}$

$\frac{10}{21}$

$\frac{9}{40}$

5 $\frac{1}{3}$

(this one is already done)

Second Row

4 $\frac{4}{9}$

5 $\frac{1}{16}$

$\frac{13}{18}$

6 $\frac{1}{4}$

11 $\frac{2}{3}$

5 0

3 years ago

Define least common multiple

AlexFokin [52]

Answer:

the smallest common multiple of two or more numbers

Step-by-step explanation:

6 0

4 years ago

Determine whether the two figures are similar. If so, give the similarity ratio of the smaller figure to the larger figure. The

geniusboy [140]

B

3x6=18
4x6=24
16x6=96
The second picture is 6 times bigger than the first so it's 1 to 6

8 0

3 years ago

Read 2 more answers

A square with side lengths of 3 cm is reflected vertically over a horizontal line of reflection that is 2 cm below the bottom ed

oksian1 [2.3K]

Answer:

a) 4 cm

b) 5 cm

c) 10 cm

Step-by-step explanation:

The side lengths of the reflected square are equal to the original, and the distance from the axis(2) also remains the same. From there, it is just addition.

Hope it helps <3

5 0

3 years ago

Solve the following matrix equations: (matrices)

Masja [62]

Step-by-step explanation:

$3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\ \\ 3X = \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \frac{1}{3} \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \begin{pmatrix} \frac{ - 3}{3} & \frac{3}{3} \\ \\ \frac{6}{3} & \frac{9}{3} \end{pmatrix}\\ \\ \huge \red{ X} = \purple{ \begin{pmatrix} - 1 &1 \\ 2 & 3 \end{pmatrix}}$

$3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\begin{pmatrix} \frac{3}{3} & \frac{0}{3} & \frac{-3}{3} \\\\ \frac{6}{3} & \frac{3}{3} & \frac{0}{3} \\\\ \frac{9}{3} & \frac{6}{3} & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1 & 0 & - 1\\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}}\\$

8 0

3 years ago