What is the missing length

Mathematics

1 answer:

steposvetlana [31]3 years ago

4 0

Answer:

25 km

Step-by-step explanation:

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mina [271]

The solution to the equation r(1 - 2cosФ) = 1 is given as x² + y² - 4x√(x² + y²) + 4x² = 1

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

In polar form:

r = √(x² + y²) and cosФ = x / √(x² + y²)

Hence:

r(1 - 2cosФ) = 1

√(x² + y²) [1 - 2(x / √(x² + y²))] = 1

√(x² + y²) - 2x = 1

Take square of both sides:

x² + y² - 4x√(x² + y²) + 4x² = 1

The solution to the equation r(1 - 2cosФ) = 1 is given as x² + y² - 4x√(x² + y²) + 4x² = 1

Find out more on equation at: brainly.com/question/2972832

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7 0

2 years ago

Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1

Lera25 [3.4K]

Assuming the function is

$f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}$

For $f(x)$ to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$
$\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)$
$\iff2=b+c$

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

$f'(x)=\begin{cases}2&\text{for }x1\end{cases}$

$\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)$
$\iff2=2b+c$

Now we solve for $b$ and $c$ :

$\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2$

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3 years ago

Help Please !!! Math

Keith_Richards [23]

The 3rd one I thinkkkkk ??

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2 years ago

find the length of each isosceles right triangle when the hypotenuse of the given measure given 5 to the square root of 6

iragen [17]

In isosceles right triangle, legs are equal. Let each leg is of x units . And the given hypotenuse is 5 square 6. We use pythagorean identity here, which is

$a^2 + b^2 = c^2$