Join all the vertices of a regular nonagon to its center. How many identical triangles are formed?

The perimeter of quadrilateral JKLM is 90 centimeters. Which is the length of side KL?

fenix001 [56]

The perimeter of an object is given by the sum of all the sides of the object. Thus the perimeter of a quadrilateral is the sum of the four sides of the quadrilateral.

Given a quadrilateral with sides: n, 2n - 12, 3n + 4, n + 7

The sum is n + 2n - 12 + 3n + 4 + n + 7 = 7n - 1 = 90

Thus, 7n = 90 + 1 = 91 and n = 91 / 7 = 13.

Therefore, the length of the side KL is n = 13 centimeters.

5 0

3 years ago

Determine the amplitude or period as requested . Amplitude of y = - 1/5 * sin x a - 1/5 1/5 b 5 d. pi/5

Brrunno [24]

Answer:

The amplitude is 1/5

Step-by-step explanation:

Given

$y = -\frac{1}{5}\sin(x)$

Required

Determine the amplitude

The general cosine function is:

$y = A\sin(k(x + c)) +d$

Where

$Amplitude = |A|$

Compare $y = A\sin(k(x + c)) +d$ and $y = -\frac{1}{5}\sin(x)$

$A = -\frac{1}{5}$

So:

$Amplitude = |-\frac{1}{5}|$

$Amplitude = \frac{1}{5}$

4 0

3 years ago

Make 'h' as the subject. Remember to rationalize the denominator while solving.

Sliva [168]

Hello there BeGuiLE !

To solve your question, we must first bring 'h' to one side of the equation. Let's bring all the terms with the variable 'h' to the left side of the equation. So,
$\large{\sf\sqrt{ 3 } h = 1.6+h}$
→ $\large{\sf\sqrt{3}h-h=1.6 }$

Now, let's combine the 2 terms containing the variable 'h'.
$\large{\sf\sqrt{3}h-h=1.6 }$
→ $\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$

Now, we need to leave the variable 'h' alone in the left side of the equation & bring (√3 + 1) to the other side of the equation. This is done in order to make 'h' as the subject of the equation. So, we'll get it as,
$\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$
→ $\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$

Now, let's rationalize it. Rationalizing is a process where we move the root from the denominator of a fraction to the numerator for easier calculation. So,
$\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1 )(\sqrt{3}+1}) }$
Now, use the algebraic identity, (a + b)(a - b) = a² - b²
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{3-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{2} }$
→ $\boxed{\large{\sf\:h=0.8(\sqrt{3}+1)}}$

Now, we've made 'h' as the subject of the equation. If you want to solve it more, then,
$\large{\sf\:h=0.8(\sqrt{3}+1)}$
Take √3 as 1.73 (approx. value up to 2 decimal places)
→ $\large{\sf\:h=0.8(1.73+1)}$
→ $\large{\sf\:h=0.8(2.73)}$
→ $\boxed{\boxed{\huge{\bf{\:h=2.184 \: (approx.)}}}}$

I hope this will help you.