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Rainbow [258]
3 years ago
9

Help with congruence

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
8 0

Answer:

H) SAS is the answer of the 9 th question

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That's a piecewise defined function and based on its definition we got the answer D.
(By combining -4≤x<2 and 2≤x≤10)
D.
- 4 ≤ x ≤ 10

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Question 16....please help me out
Varvara68 [4.7K]

Answer:

Persian-Maine Coon-American Shorthair

Step-by-step explanation:

If you look back at the question, you will see the numbers 13.65,13.07, and 13.6. So, we'll do this by digits.

The first digit of all the numbers is 1. So we'll move on. The second digit is a3, of which all numbers have in common. So we'll move on again. So now ur down to the digits 6, 0, and 6. Well, 13.07 belongs to the Persian. Then You'll see a 6, which belongs to the Maine coon. Lastly, you have another 6, which goes to the American shorthair. Correct me if i'm wrong :-)

5 0
4 years ago
How do i solve the inequality &amp; put it on a number line.|3x + 11 &gt; 2
Karo-lina-s [1.5K]

here's how you solve it

|3x+11| > 2

apply the absolute rule so you get two equations

3x+11 < -2                                    3x+11 > 2

in both equations subtract 11 from both sides

11-11 = 0    -2-11 = -13            11-11 = 0      2-11 = -9

this leaves you with

3x < -13                                        3x > -9

in both equations divide 3 on both sides

3x/3 = x     -13/3                   3x/3= x      -9/3 = -3

and in the end you get

x < -13/3  or x > -3

5 0
4 years ago
Suppose you choose a team of two people from a group of n &gt; 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
4 years ago
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