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vichka [17]
2 years ago
6

K is the midpoint of IJ complete the proof that HIK = HJK

Mathematics
1 answer:
lukranit [14]2 years ago
6 0

Answer:

um

Step-by-step explanation:

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A recreation center is offering special prices on its pool passes and gym memberships for the summer. On the first day of the of
s344n2d4d5 [400]

Answer:

In grade 8, students learned that some systems of linear equations have many solutions. This warm-up reminds students about this fact, while also prompting them to use what they learned in this unit to better understand what it means for a system to have infinitely many solutions.

The first equation shows two variables adding up to 3, so students choose a pair of values whose sum is 3. They notice that all pairs chosen are solutions to the system. Next, they try to find a strategy that can show that there are countless other pairs that also satisfy the constraints in the system. Monitor for these likely strategies:

Solving by graphing: The graphs of the two equations are the same line, so all the points on the line are solutions to the system.

Solving by substitution: The first equation can be rearranged to . Substituting for in the second equation gives or or . This equation is true no matter what is.

Solving by elimination: If we multiply the first equation by 4, rearrange the second equation to , and then subtract the second equation from the first, the result is . Subtracting from each side gives , which is true regardless of what or is.

Reasoning about equivalent equations: If we rearrange the second equation so that the variables are on the same side , we can see that this equation is a multiple of the first and are equivalent. This means they have the exact same solution set, which contains infinite possible pairs of and .

Identify students using different strategies and ask them to share their thinking with the class later.

Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).

Step-by-step explanation:

✒✒✏✂✂✏✒✏✒✂✏✂✏✒✂✒✂✂✂✏✂✒✂✂✂✒✏✏✏✒✒✏✏

7 0
3 years ago
Suppose we want to choose 6 letters, without replacement, from 15 distinct letters. (A) how many ways can this be done, if the o
Olenka [21]

Answer:

Below in bold.

Step-by-step explanation:

A.This is the number of combinations of 6 from 15

= 15C6

=  15! / (15-6)! 6!

= 5,005 ways.

B.  This is the number of permutaions of 6 from 15:

= 15! / (15-6)!

= 3,603,600 ways.

7 0
2 years ago
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