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2 years ago

6

The legs of a right triangle are 3 and 8 centimeters long, respectively. What is the length of the hypotenuse? If necessary, rou

nd your answer to two decimal places.

1 answer:

vredina [299]2 years ago

3 0

The Equation for this is A squared plus B squared equals C squared. So this would become 3 squared (9) plus 8 squared (64) equals your <span>hypotenuse squared. When you add these together you get 73. You then must find the square root of 73 which is 8.5. The length of the </span><span>hypotenuse is 8.5.</span>

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I need to simplify this

Korvikt [17]

Hi there!

$3 \sqrt{54} + 2 \sqrt{24} =$
First we split up the square root into two parts.

$3 \times \sqrt{9} \times \sqrt{6} + 2 \times \sqrt{4} \times \sqrt{6} =$
Now we calculate the value of the square roots which have an integer as a solution

$3 \times 3 \times \sqrt{6} + 2 \times 2 \times \sqrt{6} =$
Multiplying the integers gives us our next step.

$9 \sqrt{6} + 4 \sqrt{6} =$
And finally we add up the roots.

$13 \sqrt{6}$

3 0

3 years ago

Find the next three terms in the geometric sequence 2, -4, 8, -16,...

blagie [28]

Answer: 32, -64, 128, -256

8 0

3 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

If Sine (x) = two-fifths and tan(x) > 0, what is sin(2x)?

ElenaW [278]

B) $\frac{4\sqrt{21}}{25}$

Correct on Edge

4 0

3 years ago

Read 2 more answers

When comparing the functions using the equation, which conclusion can be made?

ankoles [38]

Answer:

domain of f(x) is x ≤ 0; domain of f^-1(x) is x ≥ 4

Step-by-step explanation:

The square root function will always give a positive value, so the opposite of the square root function will give non-≤positive values. That means ...

the domain of f(x) is restricted to non-positive values
the domain of the inverse function is restricted to values of x that make the root be of a non-negative number: x ≥ 4

The domain of f(x) is x ≤ 0; the domain of f^-1(x) is x ≥ 4.

8 0

3 years ago