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Paul [167]
3 years ago
14

given that 8-sqrt(18) over sqrt(2) = a + b sqrt(2) where a and b are integers find the value of a and b

Mathematics
2 answers:
Novay_Z [31]3 years ago
6 0

Answer:

a = - 3 and b = 4

Step-by-step explanation:

Given

\frac{8-\sqrt{18} }{\sqrt{2} }

Simplify \sqrt{18}

= \sqrt{9(2)} = \sqrt{9} × \sqrt{2} = 3\sqrt{2}

Thus expression can be written as

\frac{8-3\sqrt{2} }{\sqrt{2} }

Multiply numerator/denominator by \sqrt{2}

noting that \sqrt{2} × \sqrt{2} = 2

= \frac{\sqrt{2}(8-3\sqrt{2})  }{\sqrt{2}(\sqrt{2})  }

= \frac{8\sqrt{2}-6 }{2}

Dividing each term on the numerator by 2

= 4\sqrt{2} - 3

= - 3 + 4\sqrt{2}

with a = - 3 and b = 4

Delvig [45]3 years ago
6 0

Answer:

a = -3, b = 4.

Step-by-step explanation:

(8 - √18) / √2

= (8 - 3√2) √2

= 8/√2  - 3

=  8√2 / 2 - 3

= -3 + 4√2

So  a + b√2 = -3 + 4√2

Comparing coefficients we have:

a = -3 and b = 4 (answer).

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Chanasia buys a plastic bucket. The bucket weighs 460 g. The label on the bucket says made with 20% recycled plastic. How many g
Eddi Din [679]

Answer:

92

Step-by-step explanation:

20% of 460 is 92 so it would be 92

5 0
3 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

4 0
3 years ago
Can someone give me the answers for the number frame and how you did it, I’m confused :)
xxTIMURxx [149]

Answer:

16, 56, ...

Step-by-step explanation:

It looks like you have to put in each cell the multiplication of the row and the column, so the columns are 2, 7, 4 etc., and the row is always 8.

First cell: 2x8 = 16

Second cell: 7x8 = 56

and so on

It teaches you to apply the multiplication in a random sequence; you should be able to do this in your head.

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Solve for x:
astraxan [27]
The answer for this question is 8
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3 years ago
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