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sesenic [268]
3 years ago
12

Please answer this correctly. if its spam or wrong i report correct you get brainliest

Mathematics
1 answer:
kvasek [131]3 years ago
3 0
The Domain is the X values and the range is the Y values

Looking at where the 3 dots are located, the X's are -3, 0 and 2 and the Y's are 3, 0 , -2


Domain{-3,0,2} range {3,0,-2}

The answer would be A.
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Step-by-step explanation:

The radius is half the length of the diameter

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Read 2 more answers
Study the exercises in which both addends are negative. What do you notice about the sum?
Sav [38]
This question would be a lot better if you had actually included the exercises. When a question talks about other questions that you already did, you need to include those too. Otherwise we don't know what the question is talking about. Fortunately I was able to figure it out without seeing the exercises, but just for future reference, you should include that extra information.

When two addends(things you add together) are negative, adding them is a lot like adding positive numbers; you just have a negative sign on the sum. For example,
4 + 5 = 9
-4 + -5 = -9
7 0
3 years ago
Ryan runs 8 miles in 75 minutes. How many minutes does he run per mile?
victus00 [196]
First you would set up the problem as a division problem


75 \div 8

and you would get the answer of


9 minutes and 37 seconds.
8 0
3 years ago
Check answer please
Cerrena [4.2K]
The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}

\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}

Hope it helps.
5 0
3 years ago
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