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3 years ago

What is the smallest integer greater than square root of 92?

Mathematics

1 answer:

vovikov84 [41]3 years ago

5 0

sqrt(81) = 9

sqrt(100) = 100

sqrt(92) is between 9 and 10.

Answer: 10

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8x-5y= 14<br> – 5х+бу = -3

podryga [215]

Answer:

y= 34/23

x= 123/46

Step-by-step explanation:

Solve for x

8x-5y= 14

x= 14/8 +5/8y

Sub x into second equation

– 5х+бу = -3

-5(14/8 +5/8y) + 6y= -3

-29/4 +23/8y = -3

y= 34/23

Sub y into any equation and solve for x

8x=483/23

x= 123/46

4 0

2 years ago

Alice says that all the factors of 8 are even.

Rom4ik [11]

1 is not even yet it is a factor of 8. Alice is wrong.

6 0

2 years ago

1/2 = 2m what does m equal

max2010maxim [7]

1/4 time 1/4 equals a half...or you can think about it as 25 plus 25 equals 50, (1/2 of a dollar)

8 0

3 years ago

A scale drawing of a rectangular park is 5 inches wide and 7 inches long. The actual park is 280 yards long. What is the area of

Zepler [3.9K]

Answer:

56,000 yards

Step-by-step explanation:

7 inches → 280 yards

1 inch → 280 ÷ 7 = 40 yards

5 inches → 40 x 5 = 200 yards

Area = 280 x 200 = 56 000 yards

7 0

2 years ago

100 points! simplify write as a product compute

Rom4ik [11]

Answer:

a) $\sqrt{61 - 24 \sqrt{5} } = - 4 + 3 \sqrt{5}$

b) $( \sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } + {2 \sqrt{bc} } )$

c) $\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 - \sqrt{5} } = - 1$

Step-by-step explanation:

We want to simplify

$\sqrt{61 - 24 \sqrt{5} }$

Let :

$\sqrt{61 - 24 \sqrt{5} } = a - b \sqrt{5}$

Square both sides of the equation:

$(\sqrt{61 - 24 \sqrt{5} } )^{2} = ({a - b \sqrt{5} })^{2}$

Expand the RHS;

$61 - 24 \sqrt{5} = {a}^{2} - 2ab \sqrt{5} + 5 {b}^{2}$

Compare coefficients on both sides:

${a}^{2} + 5 {b}^{2} = 61 - - - (1)$

$- 24 = - 2ab \\ ab = 12 \\ b = \frac{12}{b} - - -( 2)$

Solve the equations simultaneously,

$\frac{144}{ {b}^{2} } + 5 {b}^{2} = 61$

$5 {b}^{4} - 61 {b}^{2} + 144 = 0$

Solve the quadratic equation in b²

${b}^{2} = 9 \: or \: {b}^{2} = \frac{16}{5}$

This implies that:

$b = \pm3 \: or \: b = \pm \frac{4 \sqrt{5} }{5}$

When b=-3,

$a = - 4$

Therefore

$\sqrt{61 - 24 \sqrt{5} } = - 4 + 3 \sqrt{5}$

We want to rewrite as a product:

${b}^{2} {c}^{2} - 4bc - {b}^{2} - {c}^{2} + 1$

as a product:

We rearrange to get:

${b}^{2} {c}^{2} - {b}^{2} - {c}^{2} + 1- 4bc$

We factor to get:

${b}^{2} ( {c}^{2} - 1) - ({c}^{2} - 1)- 4bc$

Factor again to get;

$( {c}^{2} - 1) ({b}^{2} - 1)- 4bc$

We rewrite as difference of two squares:

$(\sqrt{( {c}^{2} - 1) ({b}^{2} - 1) })^{2} - ( {2 \sqrt{bc} })^{2}$

We factor the difference of square further to get;

$( \sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2} - 1) ({b}^{2} - 1) } + {2 \sqrt{bc} } )$

c) We want to compute:

$\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 - \sqrt{5} }$

Let the numerator,

$\sqrt{9 - 4 \sqrt{5} } = a - b \sqrt{5}$

Square both sides of the equation;

$9 - 4 \sqrt{5} = {a}^{2} - 2ab \sqrt{5} + 5 {b}^{2}$

Compare coefficients in both equations;

${a}^{2} + 5 {b}^{2} = 9 - - - (1)$

and

$- 2ab = - 4 \\ ab = 2 \\ a = \frac{2}{b} - - - - (2)$

Put equation (2) in (1) and solve;

$\frac{4}{ {b}^{2} } + 5 {b}^{2} = 9$

$5 {b}^{4} - 9 {b}^{2} + 4 = 0$

$b = \pm1$

When b=-1, a=-2

This means that:

$\sqrt{9 - 4 \sqrt{5} } = - 2 + \sqrt{5}$

This implies that:

$\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 - \sqrt{5} } = \frac{ - 2 + \sqrt{5} }{2 - \sqrt{5} } = \frac{ - (2 - \sqrt{5)} }{2 - \sqrt{5} } = - 1$

3 0

2 years ago

Read 2 more answers