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Andrej [43]
3 years ago
14

Which congruency statements are true?

Mathematics
2 answers:
MArishka [77]3 years ago
7 0

Answer:

QH≅MF, MN≅QR, M≅Q,

djyliett [7]3 years ago
4 0

Answer: 1st answer, 2nd answer, and 4th answer

Step-by-step explanation:

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The graph below shows the numbers of visitors at a museum over six days.
Bingel [31]

Answer:

(a) 305 visitors (b): Saturday to Sunday

5 0
3 years ago
Help me please hurry :)
Fudgin [204]

Answer:

1. f>30

2. s≥140

Hope This Helps!!!

5 0
3 years ago
100 POINTS 10 QUESTIONS PLEASE HELP PICTURES BELOW PLEASE SPECIFY WHICH QUESTION YOU ARE ANSWERING
Semenov [28]

Answer:

The answers are \frac{2}{5}, \frac{1}{4}, \frac{1}{10}, \frac{5}{2}, \frac{5}{8}, \frac{4}{1}, \frac{8}{5}, and \frac{10}{1}.

Step-by-step explanation:

Proportions are fractions that can be made by using the given numbers, which in this case are 2, 5, 8, and 20. Let's pair each one with the other three and then simplify if possible.

First, let's begin with 2:

\frac{2}{5} = \frac{2}{5}

\frac{2}{8} = \frac{1}{4}

\frac{2}{20} = \frac{1}{10}

Then, let's do 5:

\frac{5}{2} = \frac{5}{2}

\frac{5}{8} = \frac{5}{8}

\frac{5}{20} = \frac{1}{4}

Note that we already have \frac{1}{4}, so we do not need to include an additional one.

Now, let us do 8:

\frac{8}{2} = \frac{4}{1}

\frac{8}{5} = \frac{8}{5}

\frac{8}{20} = \frac{2}{5}

See how we already have \frac{2}{5}, so we won't have to include that as well.

Finally, let's do 20:

\frac{20}{2} = \frac{10}{1}

\frac{20}{5} = \frac{4}{1}

\frac{20}{8} = \frac{5}{2}

Now see that we already have both \frac{4}{1} and \frac{5}{2}, so we won't have to include both of them, as they are both extras.

Hence, the answers are \frac{2}{5}, \frac{1}{4}, \frac{1}{10}, \frac{5}{2}, \frac{5}{8}, \frac{4}{1}, \frac{8}{5}, and \frac{10}{1}.

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
8 0
3 years ago
Read 2 more answers
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
Homework please help its over due​
xxMikexx [17]

Step-by-step explanation:

TRIANGLE 1

c = 2√5

= 2 × √5

= √4 × √5

= √20

a = 2

b = ?

b² = c² - a²

= √20² - 2²

= 20² - 4

= 400 - 4

= 396

b = √396

= √36 × √11

= 6 × √11

= 6 √11

TRIAGLE 2

a = 12

b = 6√11

= √396

c = ?

c² = a² + b²

= 12² + √396

= 144 + √396²

= 144 + 396

c = √540

= √36 × √15

= 6 × √15

= 6 √15

Sorry if i wrong. Im beginner in USA server

4 0
3 years ago
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