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2 years ago

BRAINLIEST PLS HELP!!!

Mathematics

1 answer:

Alik [6]2 years ago

8 0

The answer is a......

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HELP me plzzzz i need to finish this

trapecia [35]

Answer:

725.95

Step-by-step explanation:

If you multiply 129.95*5. You get 649.75, but you have to add 76.20 from the delivery and you get 725.95

Hope this helps =)

4 0

2 years ago

11. 4(x+1)=5(x-3)<br> help please

soldier1979 [14.2K]

$4(x + 1) = 5(x - 3) \\ 4x + 4 = 5x - 15 \\ 5x - 4x = 4 + 15 \\ x = 19$

4 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

3 years ago

Read 2 more answers

Can anyone help me find the area of each polygon with explanation ?

ivanzaharov [21]

Answer:

ok so basically since it is all covered in just count the squares that are shaded

8 0

2 years ago

Read 2 more answers

The number of miles to the next service station and the number of kilometers is linear or not

Marysya12 [62]

Yes. Because miles and kilometers are units of distance measurements. They have a fixed ratio, hence, there will always be a constant linear relationship between them. Or a constant slope. One mile is approximately 1.61 kilometers. So, just multiply any measurement of mile with 1.61 to get the distance in km.

8 0

3 years ago