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3 years ago

Sq root of 2 minus 1 ÷ sq root of 2 plus 1

Mathematics

1 answer:

elena-s [515]3 years ago

7 0

√2 - 1 (√2 - 1)(√2 - 1)

--------- = ---------------------

√2 + 1 (√2 + 1)(√2 - 1)

2 + 1 - 2√2

= ---------------

2 - 1

= 3 - 2√2 Answer

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517 37/50 + 312 3/100

fgiga [73]

Answer: 829 3/4 or 829 75/100

Step-by-step explanation:

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3 years ago

A variable is a placeholder for an unknown value

frosja888 [35]

Answer:

Im going to guess that this is a true or false question so..

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3 years ago

How would i graph this equation y=3x+1

Kipish [7]

Answer:

(0,1)(1,4)

Step-by-step explanation:

Put in any value of x to find y, then you have a coordinate, then plot it

lets say i pick 0 for x

y = 3(0) + 1

y = 1

(0,1)

lets say i also pick 1 for x

y = 3(1)+1

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3 years ago

Jeff is baking a cake. The recipe says that he has to mix 32 grams of vanilla powder to the flour. Jeff knows that 1 cup of that

Sedaia [141]

Answer:

Jeff went over by 53.33 grams or $53\frac{1}{3}$ grams.

Step-by-step explanation:

Jeff is baking a cake. The recipe says that he has to mix 32 grams of vanilla powder to the flour.

Jeff knows that 1 cup of that particular vanilla powder has a mass of 128 grams.

He added $\frac{2}{3}$ of a cup of vanilla powder to the flour.

So, he added a mass of $\frac{2}{3} \times128= 85.33$ grams of vanilla.

As the recipe says, it needs 32 grams of vanilla so Jeff went over by $85.33-32=53.33$ grams.

Jeff went over by 53.33 grams or $53\frac{1}{3}$ grams.

8 0

3 years ago

Need answer to 4 and 5

algol13

Answer:

Substitute n =1, 2, 3, 4, 5 to get the first five terms.

Step-by-step explanation:

(4). $g_n = 2 . 3^{n - 1}$

To find the first term, substitute n = 1.

Therefore, $g_1 = 2 . 3^{1 - 1} = 2 . 3^{0} = 2 . 1$ = 2

$g_2 = 2 . 3^{2 - 1} = 2 . 3$ = 6

$g_3 = 2 . 3^{3 - 1} = 2 . 3^{2} = 2 . 9$ = 18

$g_4 = 2 . 3^{4 - 1} = 2 . 3^{3} = 2 . 27$ = 54

$g_5 = 2 . 3^{5 - 1} = 2 . 3^{4} = 2 . 81$ = 168

Therefore the first five terms of the sequence are: 2, 6, 18, 54, 168.

(5). $t_n = \frac{2}{3}t_{n - 1}$

$t_2 = \frac{2}{3} t_{2 - 1} = \frac{2}{3}t_1 = \frac{2}{3}6$ = 4

$t_3 = \frac{2}{3} t_2 = \frac{2}{3}. 4 = \frac{8}{3}$

$t_4 = \frac{2}{3}t_3 = \frac{2}{3}\frac{8}{3} = \frac{16}{9}$

$t_5 = \frac{2}{3}t_4 = \frac{2}{3}\frac{16}{9} = \frac{32}{27}$

Therefore the first five terms in the sequence are: 6, 4, 8/3, 16/9, 32/27.

8 0

3 years ago