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Sauron [17]
3 years ago
14

Find the distance between the points (3, -5) and (-6, -5).

Mathematics
1 answer:
Sophie [7]3 years ago
8 0

ANSWER

9

EXPLANATION

We want to find the distance between the points (3, -5) and (-6, -5).

The given points have the same y-coordinates .

This means it is a horizontal line.

We use the absolute value method to find the distance between the two points.

We find the absolute value of the distance between the x-values.

The distance between the two points is

|3--6|=|3+6|=|9|=9

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Pls help with this question
tatyana61 [14]

Answer:

-9, -2, -1.0001, -6, -1.1, -4, and -1.01

Step-by-step explanation:

Anything to the left of -1 on the line means that the number is less than -1.

5 0
3 years ago
PLZZZ HELP FAST What is the product of Three-fourths and Negative StartFraction 6 over 7 EndFraction? Negative StartFraction 7 o
8_murik_8 [283]

Answer:

-9/14 so the answer is B

Step-by-step explanation:

Hope this helps

8 0
2 years ago
When addressing an envelope for delivery in the United States or Canada, the zip code should appear
levacccp [35]
ZIP codes are codes utilized for the purpose of postal and adhering to the postal system. This contributes to faster process of delivering to and from the send and receiver of the mails. ZIP codes do not just apply on countries worldwide, but this also applies to the places inside these countries. Provinces and cities and have their own ZIP codes assigned to them in accordance to the ZIP code assigned to the country. This is why most mails should have a ZIP code attached to the mail and to the letter when being delivered whether it is locally delivered or internationally delivered. When addressing an envelope for delivery in the United States or Canada, the ZIP code should appear on the same line as the city and state. So the answer to the question above would be B.
6 0
3 years ago
Let g' be the group of real matricies of the form [1 x 0 1]. Is the map that sends x to this matrix an isomorphism?
aliina [53]

Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component x. So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.

However, let's prove in a more formal way that

\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]

is an isomorphism.

First of all, it is injective: suppose x \neq y. Then, you trivially have \phi(x) \neq \phi(y), because they are two different matrices:

\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]

Secondly, it is trivially surjective: the matrix

\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]

is clearly the image of the real number x.

Finally, \phi and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have

\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)

8 0
3 years ago
Round 3487 to the nearest thousand​
dlinn [17]

Answer:

3000

Step-by-step explanation:

the 4 next to the 3 means that you will be rounding down.

6 0
3 years ago
Read 2 more answers
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