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insens350 [35]
3 years ago
11

Are these congruent by SSS, SAS, AAS, ASA?

Mathematics
1 answer:
faust18 [17]3 years ago
7 0

Answer:

Step-by-step explanation:

ASA

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What is the total number of digits used when the first 2002 positive even integers are written?
Luda [366]

The total number of digits used when the  first 2002 positive even integers are written is 4 digits

<h3>How to find the number of sequence?</h3>

Using arithmetic sequence,

nth term = a + (n - 1)d

where

  • a = first term
  • d = common difference
  • n = number of terms

a = 2

d = 2

nth term = 2 + (2002 - 1)2

nth term = 2 + 4002

nth term = 4004

learn more on sequence here: brainly.com/question/15456604

#SPJ1

7 0
2 years ago
Escribe una ecuación para una linea que pase por los puntos (3,5) y (6,3). Responder con trabajo de apoyo​
Blababa [14]

Answer:

y = -\frac{2}{3}x + 7

Step-by-step explanation:

Ambos puntos (3, 5) y (6, 3) están en esta línea.

8 0
3 years ago
Evaluate.<br> 2(4 what is the correct answer?
Kruka [31]
The Answer Would Be 16

How I Found The Answer:
2 x 2 x 2 x 2 = 16
3 0
3 years ago
Read 2 more answers
Helppp plzzz idk I need help​
spin [16.1K]

Answer:

<em>T</em><em>h</em><em>e</em><em> </em><em>c</em><em>o</em><em>r</em><em>r</em><em>e</em><em>c</em><em>t</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>i</em><em>s</em><em> </em><em>y</em><em>.</em>

Step-by-step explanation:

<em>h</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>w</em><em>o</em><em>r</em><em>k</em><em>s</em><em> </em><em>o</em><em>u</em><em>t</em><em> </em><em>!</em><em>!</em>

3 0
3 years ago
How many strings can be formed by ordering the letters MISSISSIPPI which
Mazyrski [523]

Answer:

<em>1680 is the answer.</em>

Step-by-step explanation:

Here, we have 11 letters in the word <em>MISSISSIPPI.</em>

Repetition of letters:

M - 1 time

I - 4 times

S - 4 times

P - 2 times

As per question statement, we need a substring MISS in the resultant strings.

So, we need to treat MISS as one unit so that <em>MISS always comes together in all the strings</em>.

The resultant strings will look like:

<em>xxxxMISSxxx</em>

<em>xxMISSxxxxx</em>

<em>and so on.</em>

After we treat <em>MISS </em>as one unit, total letters = 8

Repetition of letters:

MISS - 1 time

I - 3 times

S - 2 times

P - 2 times

The formula for combination of letters with total of <em>n </em>letters:

\dfrac{n!}{p!q!r!}

where <em>p, q</em> and <em>r</em> are the number of times other letters are getting repeated.

<em>p = 3</em>

<em>q = 2</em>

<em>r = 2</em>

<em></em>

So, required number of strings that contain MISS as substring:

\dfrac{8!}{3!2!2!}\\\Rightarrow \dfrac{40320}{6\times 2 \times 2}\\\Rightarrow 1680

So, <em>1680 is the answer.</em>

3 0
3 years ago
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