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bekas [8.4K]
3 years ago
5

Select the reason why these triangles are similar. If they are not, select “not similar”

Mathematics
1 answer:
larisa [96]3 years ago
8 0

Answer:

B. SAS

Step-by-step explanation:

3/1= 2.25/075

The 2 sides are congruent and the angle between them is common

B. SAS is the answer

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What is the mean absolute deviation for the set of data below?<br><br> 2, 8, 10, 16
Evgesh-ka [11]

Solution: We are given below data:

2, 8, 10, 16

Now to find the mean deviation, we use the below formula:

MD= \frac{\sum|X-\bar{X}|}{N}

Where:

\sum,  represents the summation

X, represents the observation.

\bar{X}, represents the mean

N represents the number of observation.

\bar{X}= \frac{\sum{X}}{N}= \frac{2+8+10+16}{4}=9

\sum |X-\bar{X}| = |2-9| + |8-9| +|10-9| +|16-9|=7+1+1+7=16

Therefore, the mean deviation is:

MD= \frac{16}{4}

        = 4

5 0
4 years ago
Read 2 more answers
Solve for x: -1/2(-4+2x) = -6 - 4x
antiseptic1488 [7]
-\dfrac{1}{2}(-4+2x)=-6-4x\ \ \ \ \ |multiply\ both\ sides\ by\ (-2)\\\\-4+2x=12+8x\ \ \ \ |add\ 4\ to\ both\ sides\\\\2x=16+8x\ \ \ \ \ |subtract\ 8x\ from\ both\ sides\\\\-6x=16\ \ \ \ \ |divide\ both\ sides\ by\ (-6)\\\\x=-\dfrac{16}{6}\\\\x=-\dfrac{8}{3}\\\\\boxed{x=-2\frac{2}{3}}
5 0
3 years ago
Given: F3 = 5 k + 515 g F4 = 945k - 713 g ; find F20
Alisiya [41]

9514 1404 393

Answer:

  F20 = 15985k -20361g

Step-by-step explanation:

If we assume F3 and F4 are the third and fourth terms of an arithmetic sequence, the common difference is ...

  d = F4 -F3 = (945k -713g) -(5k +515g)

  d = 940k -1228g

Then the 20th term is ...

  F20 = F3 +(20 -3)d = (5k +515g) +17(940k -1228g)

  F20 = 15985k -20361g

8 0
3 years ago
What's the product of 3 2/3 and 14 2/5 ?<br> A. 52 4/5<br> B. 52 4/15<br> C. 54<br> D. 42 4/5
wlad13 [49]

Answer:

52 4/5

Have a nice day! :)

4 0
3 years ago
35
Roman55 [17]

\neq  \lim_{n \to \infty} a_n \pi x_{123} \alphaAnswer:

Step-by-step explanation:

There are 3 boxes. The formula for volume is lxhxw, (length x height x width)

6 0
2 years ago
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