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BigorU [14]
3 years ago
15

Please Help!!!

Mathematics
1 answer:
vaieri [72.5K]3 years ago
7 0

I DON'T UNDERSTAND MATH ANYMORE!!!!!!!!

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Kevin is 333 years older than Daniel. Two years ago, Kevin was 444 times as old as Daniel.
tia_tia [17]

Answer: the equations are

k = d + 33

k - 2 = 4(d-2)

Step-by-step explanation:

Let the present age of Kevin be represented by k

Let the present age of Daniel be represented by d

Kevin is 33 years older than Daniel. This means that the expression for their current age is

k = d + 33

Two years ago, Kevin was 4 times as old as Daniel. This means that 2 will be subtracted from their present ages to depict two years ago. Therefore, Two years ago for Daniel will be d-2

Two years ago for Kevin will be k - 2

Remember that Kevin was 4 times as old as Daniel two years ago, it becomes

k - 2 = 4(d-2)

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3 years ago
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What is 4 as a ratio
Zepler [3.9K]

Answer:

4:1

Step-by-step explanation:

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3 years ago
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Solve the following matrix equations: (matrices)
Masja [62]

Step-by-step explanation:

a)

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}}

b)

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
What are the vertical angles
wariber [46]
It’s the bottom left answer because the angles are opposite of each other. They are also congruent. I hope this helps!
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2×3 squared<br><img src="https://tex.z-dn.net/?f=2%20%5Ctimes%203%20%7B%3F%7D%5E%7B2%7D%20" id="TexFormula1" title="2 \times 3 {
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2 x 3^2
2 x 9
18

18 is the answer
3 0
2 years ago
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