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dezoksy [38]
2 years ago
12

Venn diagrams in maths

Mathematics
1 answer:
OleMash [197]2 years ago
5 0

Answer:

Step-by-step explanation:

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can someone help out ?? question is "The cost of Caroline’s cell phone plan is shown below in the table. What is the equation th
butalik [34]

Answer:

y = 32x

We notice that we multiply the number of months by 32

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3 years ago
329,444,000,777,234 in words​
maxonik [38]

Step-by-step explanation:

Three hundred and twenty nine trillion four hundred and fourty four billion seven hundred and seventy seven thousand two hundred and thrity four

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3 years ago
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The bearing of a plane from the airport is 65.
Ulleksa [173]

Answer:

The bearing of the airport from the plane is 245°

Step-by-step explanation:

The bearing of a point from a location is given by the angle in degrees of rotation from the Northern direction of the location to the direction of the location of the point

The given bearing of the plane from the airport = 65°

Therefore, the direction of the plane from the Northern direction of the airport = 65°

From the Northern direction of the plane, the bearing of the airport = 245°

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3 years ago
If the edge length of a cube is 11 feet, what is the correct way to write the expression to represent the volume of the cube in
Keith_Richards [23]
11^3 would be exponential form.

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Hope this helps!
5 0
3 years ago
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Please help me. I don’t understand matrices
emmasim [6.3K]

Answer:

A. \orange {\begin{bmatrix} 4 & 11 \\ 3 & 0 \end{bmatrix}}

Step-by-step explanation:

A= \begin{bmatrix} 5 & - 8 \\ 3 & 0\end{bmatrix} \: and \: B= \begin{bmatrix} 2 & - 9 \\ 1 & 0\end{bmatrix}

\therefore 2A= 2\begin{bmatrix} 5 & - 8 \\ 3 & 0\end{bmatrix} \: and \:  \: 3B= 3\begin{bmatrix} 2 & - 9 \\ 1 & 0\end{bmatrix}

\therefore 2A= \begin{bmatrix} 10 & - 16 \\ 6 & 0 \end{bmatrix} \: and \: \:  3B= \begin{bmatrix} 6 & - 27 \\ 3 & 0\end{bmatrix}

\therefore 2A-3B= \begin{bmatrix} 10 & - 16 \\ 6 & 0 \end{bmatrix} - \begin{bmatrix} 6 & - 27 \\ 3 & 0 \end{bmatrix}

\therefore 2A-3B= \begin{bmatrix} 10-6 & - 16-(-27) \\ 6-3 & 0-0 \end{bmatrix}

\therefore 2A-3B= \begin{bmatrix} 4 & - 16+27 \\ 3 & 0 \end{bmatrix}

\huge \red {\therefore 2A-3B= \begin{bmatrix} 4 & 11 \\ 3 & 0 \end{bmatrix}}

3 0
2 years ago
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