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Ymorist [56]
3 years ago
5

Find the area A of the polygon with the given vertices.

Mathematics
1 answer:
Svetllana [295]3 years ago
8 0

Answer:

9

Step-by-step explanation:

no

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I need help <br> 5=6(1q-5)-19
Rudiy27

ok distribute the 6 by multiplying it by 1q and -5
5=6q-30-19
Now subtract 30-19
5=6q-11
add 11 to both sides
16=6q
divide both sides by 6
q= 16/6
simplify
16/2=8, 6/2=3
q=8/3
7 0
3 years ago
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A recipe for a loaf of bread calls for 2/3 of a cup of flour. If Milo used 12 cups of flour, how many loafs of bread did he prep
Aloiza [94]
The answer is:

1/2 cups of flour can make 1 loaf.

1 cup would make =   1 /(1/2) = 2 loaves.

12 cups of flour would then make =  12*2 = 24

= 24 loaves.

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7 0
3 years ago
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Julie is a cashier. She can ring up 12 customers in 9 mintesutes. At the rate, how many minutes does it take for her to ring up
oksano4ka [1.4K]

Answer:

5.2

Step-by-step explanation:

ok so we need to find the unit rate( how many customers per minute) and thats 12 divided by 9 which is 1.3 minutes per customer so now we need to find what 1.3 minutes is x4 customers. This brings us to 5.2.

6 0
2 years ago
Simplify 2b –7+3b+4 –b+3.
madam [21]

Answer:

4

7 0
3 years ago
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How many different combinations are possible if each lock contains the numbers 0 to 39, and each combination contains three dist
Georgia [21]
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
8 0
3 years ago
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