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morpeh [17]
3 years ago
8

What is the simplest form of the radical expression? √2+√5/√2-√5

Mathematics
1 answer:
kicyunya [14]3 years ago
4 0

here u go...........

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Omg i keep mixing up the questions I JUST NEED HELP ON PART A THATS IT
ra1l [238]

Answer:

<u>1/557</u>

Step-by-step explanation:

The <u>scale factor</u> on a map is the ratio of one unit to the distance that unit measures!

8 0
1 year ago
You wish to mentally estimate the total cost of items that have the following prices: $1.85, $.98, $3.49, $9.78, and $6.18. Roun
Mkey [24]
22.50$ with rounding to the nearest half dollar.
7 0
3 years ago
Read 2 more answers
Find an expression which represents the difference when (2x − 6) is subtracted
Sloan [31]

Answer:

-2x + 8

Step-by-step explanation:

(2x - 6) is subtracted from (7 - 5).

First term that into an equation.

Because our first term is being subtracted from our second one, our equation is written as:

(7 - 5) - (2x - 6).

Simplify the first term by subtracting:

7 - 5 = 2

Now simplify the second term by multiplying 2x - 6 by -1 to get them out of the parenthesis. We do this because there is a negative sign in front of it.

-1 (2x - 6)

-2x + 6

Now our equation is:

2 - 2x + 6

Combine like terms:

2 + 6 = 8

So -2x + 8

(2x - 6) subtracted from (7 - 5) is -2x+8

4 0
1 year ago
36 creates of apples. each creat has 25 kgs of apples. they sell 486,235 games of apples. how many grams are left?
valentinak56 [21]

The grams of apples left is 413765 grams

<h3>How to determine the value</h3>

From the information given, we have that:

  • There are 36 crates of apples
  • Each contains 25 kilograms of apples
  • 486,235 grams of apples were sold

If we have 36 crates, let's determine the the total grams of apples

1 crate = 25 kilograms = 25000 grams

36 crates = x

x = 25000 × 36

x = 900000 grams

The total grams of apple is  900000 grams

The grams left = Total - sold =  900000 - 486,235 = 413765 grams

Thus, the grams of apples left is 413765 grams

Learn more about word problems here:

brainly.com/question/1781657

#SPJ4

8 0
2 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
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