A polygon is shown on the graph

Mathematics

2 answers:

boyakko [2]3 years ago

8 0

(x+5,y-4) :The Answer

Grace [21]3 years ago

6 0

Answer:

A’(-1,1), B’(-1,-2), C’(3,-2), D’(3,2)

Step-by-step explanation:

Basically it's just (x+5,y-4)

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How to solve theorems

Alex73 [517]

Use the Pythagorean theorem to calculate the value of X. Round your answer to the nearest hundredth. Remember our steps for how to use this theorem. This problems is like example 2 because we are solving for one of the legs. Identify the legs of the right triangle.

hope that helped

5 0

4 years ago

Round 141 decimal point 999 to the nearest tenth hundredth ten and hundred

eduard

141.999 rounded to the nearest
tenth = 142
hundreth =142
ten = 140
hundred = 100

3 0

3 years ago

Help pls I'll give 25 points

RideAnS [48]

Answer:

$\text{C. }60$

Step-by-step explanation:

Question from image:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the x marks where it cannot be.

$\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}$

Let's start with the first position:

$\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in $3!=6$ ways. Therefore, there are $4\cdot 6=\boxed{24}$ possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

$\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in $3!=6$ ways. Therefore, there are $3\cdot 6=\boxed{18}$ possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are $3!=6$ ways to rearrange the remaining 3 digits for a total of $2\cdot 6=\boxed{12}$ possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still $3!=6$ ways to rearrange the remaining three numbers. Therefore, there are $1\cdot 6=\boxed{6}$ possible numbers when 1 is the fourth digit of the number.

Thus, there are $24+18+12+6=\boxed{60}$ numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.