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AleksandrR [38]
3 years ago
10

10 points. Please don't give a answer that doesn't make sense.

Mathematics
1 answer:
katovenus [111]3 years ago
5 0

Answer:

ok

Step-by-step explanation:

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A pair of equations is shown below: y = 8x − 9 y = 4x − 1 Part A: Explain how you will solve the pair of equations by substituti
kicyunya [14]

Answer:

for substitution, make one of the alphabet (x or y) a subject from one of the equation and substitute it in the other equation.

for elimination, name the equations as 1 and 2, and try to find a possible equation 3 then substitute equation 3 into equation 1 or 2 and you'll finally get your answer

should I solve it?

5 0
3 years ago
If a gambler rolls two dice amd gets the sum of 5 he wins $30. If he gets a sum of 3 he wins $50. The cost of playing is $15. Wh
larisa [96]

Answer:

C) -$8.89

Step-by-step explanation:

If he rolls two dice and gets the sum of 5, he wins $30.

If he gets the sum of 3, he wins $50

While rolling two dice, there are 36 possible outcomes (6^2) = 36

There are four possible ways (1,4 3,2 2,3 4,1)of obtaining the sum of 5 out of 36 possible outcomes.

The probability of expecting the sum of 5= 4/36

= 1/9

This means that on every try we expect $30/9

There are two possible ways (1,2 2 2,1)of obtaining the sum of 3 out of 36 possible outcomes.

The probability of expecting the sum of 5= 2/36

= 1/18

This means that on every try we expect $50/18

For every 9 rolls, you expect to get $30 and on every 18, you expect to get $50. On any particular roll, you should expect to gey

$30/9 + $50/18

= (60+50)/18

= 110/18

= $6.11

Given that each game he is playing cost $15. The overall expectations is to lose (15- 6.11) = $8.89 each time he plays.

Answer = -$8.89

8 0
4 years ago
Simplify<br>6(2 - 7v) - 4(-6v + 7)​
FrozenT [24]

6(2 - 7v) - 4(-6v + 7)​

multiply the first bracket by 6

(6)(2)=-12

(6)(-7v)=-42v

multiply the second bracket by -4

(-4)(-6v)=-24v

(-4)(7)=-28

12-42v+24v-28

-16-18v

Answer:

-16-18v or -18v-16

6 0
4 years ago
Read 2 more answers
Let A ⊆ B ⊆ C be rings. Suppose C is a finitely generated A-module. Does it follow that B is a finitely generated A-module?
Lunna [17]

Answer:

Let A ⊆ B ⊆ C be rings. If C is a finitely generated A-module. Then B is a finitely generated A-module.

Step-by-step explanation:

Draw a ring and call it A, then draw another circle with a longer radius from the same centre of A and call it B, then draw another from the same centres of A and B, but with the longest radius and call it C.

Then, when you say A ⊆ B ⊆ C, this means that A is a subset of and equal to B which is a is a subset of and equal to C.

Meaning:

1. A is in B and B is in C.

2. The values in A are the only values in B. i.e

If A = {2,4,6} then B = {2,4,6}

3. The values of ring B are the only values in ring C. i.e. if B = {2,4,6} then C = {2,4,6}.

4. There is no more values in B that is not in A.

5. There are no more values in C that is not in B.

Since they are subsets of each other defined by ⊆, which makes the subset exactly the same as the host set or superset.

So the same rule that applies to C will apply to B

A finitely generated module is a module that has a finite generating set. A finitely generated module over a ring A may also be called a finite A-module.

8 0
3 years ago
PLEASE HELP FAST!! 30 POINTS
Ainat [17]

Answer:

-2.3, .35, 4/11 ,9/10, sqrt(7)

Step-by-step explanation:

put them all in decimal form

.35

9/10 = .9

sqrt(7) = 2.64575

-2.3

4/11 =.363636........

smallest to largest

-2.3, .35, .363636........,.9, 2.64575


in terms of the original numbers

-2.3, .35, 4/11 ,9/10, sqrt(7)


6 0
3 years ago
Read 2 more answers
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