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azamat
2 years ago
8

Does anyone know the answer to this I'm k12 btw​

Mathematics
2 answers:
VMariaS [17]2 years ago
7 0
2nd and 4th are the answers
kirill115 [55]2 years ago
7 0
The Second one and the fourth one
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Create a word problem on Algebraic Expressions and Measurement (grade 10)​
mrs_skeptik [129]

Frank can build a fence in twice the time it would take Sandy. Working together, they can build it in 7 hours. How long will it take each of them to do it alone?

Answer

If Frank and Sandy can build the fence in 7 hours, they must be building

1

7

of the fence every hour.

Now, let the amount of time it takes Sandy be

x

hours so that Frank takes

2

x

hours. Sandy can build

1

x

of the fence every hour and Frank can build

1

2

x

of the fence every hour.

We now have the following equation to solve.

1

x

+

1

2

x

=

1

7

2

+

1

2

x

=

1

7

21

=

2

x

x

=

21

2

=

10.50

Thus, Sandy takes

10.50

hours and Frank takes

21

hours.

6 0
3 years ago
B
denis23 [38]

Answer:

C. 5units

Step-by-step explanation:

7 0
3 years ago
Can someone help me ? Please thank you
Tom [10]
Correct answer is the 4th choice 16!!!

Hope this helps!!!
7 0
3 years ago
The mass of an adult North Pacific right whale is about 70,000 kg, or 7 x 104
olasank [31]

Answer:

Answer B: 1*10^2 times

Step-by-step explanation:

To find how many times more massive is the whale relative o the polar bear, we just need to find the quotient of one over the other:

\frac{7*10^4}{7*10^2} = \frac{7*10^2*10^2}{7*10^2}=1*10^2 times

Therefore our answer agrees with the one shown as answer B.

6 0
3 years ago
Read 2 more answers
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
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