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djyliett [7]
3 years ago
13

What is the midpoint of the segment with endpoints (1,9)and(-3,5)

Mathematics
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:

(-1,7)

Step-by-step explanation:

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A figure was translated 4 units down and 3 to the left then reflected over the y-axis. The figures are
Mumz [18]

Answer:

Congruent and similar

Step-by-step explanation:

because you did not change anything about the shape that shape is still the same size and shape just in a different spot

4 0
3 years ago
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Please answer the following 4 questions
Lilit [14]
1. The correct answer should be A
2. The answer should be C
3. I think the answer is D
4. The answer should be D


Hope this helps :)
7 0
3 years ago
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Mila and samuel had an equal amount of money for shopping. Mila spent $42 and Samuel spent $36. After that, Mila had 1/2 of what
koban [17]

Answer:

Mila had $48 initially

Step-by-step explanation:

Let the amount of money they had be x

Mila spent $42; what is left would be x-42

Samuel spent $36: what is left would be x-36

So, what Mila had left is half of what Samuel has; that would be;

x-42 = 1/2(x-36)

2(x-42) = x -36

2x-84 = x -36

2x - x = 84 -36

x = $48

6 0
3 years ago
In a school with 117 seventh-grade students and about 29 students per class, the number of seventh-grade classes is about
notka56 [123]

Answer - 4 classes

117 ÷ 29 = 4.03...

That is the answer. hope this helps.

5 0
3 years ago
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Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
2 years ago
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